CHAPTER 4
20
40
PROBLEM 4.1
20
20
A
Knowing that the couple shown acts in a vertical plane, determine
the stress at (a) point A, (b) point B.
M 5 15 kN · m
80
20
B
Dimensions in mm
SOLUTION
For rectangle:
I
1 3
bh
12
Outside rectangle:
I1
1
(80)(120)3
12
I1
11.52 106 mm 4
I2
1
(40)(80)3
12
I2
1.70667 106 mm 4
Cutout:
Section:
(a)
yA
40 mm
I
I1
I2
11.52 10
6
m4
1.70667 10
9.81333 10
0.040 m
6
A
6
m4
m4
My A
I
(15 103 )(0.040)
9.81333 10 6
61.6 106 Pa
61.6 MPa
A
(b)
yB
60 mm
0.060 m
B
MyB
I
(15 103 )( 0.060)
9.81333 10 6
91.7 106 Pa
B
91.7 MPa
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447
2 in. 2 in. 2 in.
PROBLEM 4.2
M ! 25 kip · in.
A
Knowing that the couple shown acts in a vertical plane, determine
the stress at (a) point A, (b) point B.
2 in.
B
1.5 in.
2 in.
SOLUTION
I
For rectangle:
1 3
bh
12
For cross sectional area:
I
I1
I2
I3
1
(2)(1.5)3
12
1
(2)(5.5)3
12
1
(2)(1.5)3
12
(a)
yA
2.75 in.
A
My A
I
(25)(2.75)
28.854
(b)
yB
0.75 in.
B
MyB
I
(25)(0.75)
28.854
28.854 in 4
A
B
2.38 ksi
0.650 ksi
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448
200 mm
PROBLEM 4.3
12 mm
y
C
M
x
Using an allowable stress of 155 MPa, determine the largest bending
moment M that can be applied to the wide-flange beam shown. Neglect the
effect of fillets.
220 mm
8 mm
12 mm
SOLUTION
Moment of inertia about x-axis:
I1
1
(200)(12)3
12
(200)(12)(104)2
25.9872 106 mm 4
I2
1
(8)(196)3
12
I3
I1
25.9872 106 mm 4
I
I1
I2
M
Mx
I3
Mc
with c
I
I
with
c
5.0197 106 mm 4
56.944 106 mm 4
1
(220)
2
110 mm
56.944 10
6
m4
0.110 m
155 106 Pa
(56.944 10 6 )(155 106 )
0.110
80.2 103 N m
Mx
80.2 kN m
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449
200 mm
PROBLEM 4.4
12 mm
y
Solve Prob. 4.3, assuming that the wide-flange beam is bent about the y axis
by a couple of moment My.
C
M
x
220 mm
PROBLEM 4.3. Using an allowable stress of 155 MPa, determine the
largest bending moment M that can be applied to the wide-flange beam
shown. Neglect the effect of fillets.
8 mm
12 mm
SOLUTION
Moment of inertia about y axis:
I1
I2
1
(12)(200)3 8 106 mm 4
12
1
(196)(8)3 8.3627 103 mm 4
12
I3
I1
8 106 mm 4
I
I1
I2
My
My
I3
Mc
with c
I
I
with
c
16.0084 106 mm 4
1
(200)
2
100 mm
16.0084 10
6
m4
0.100 m
155 106 Pa
(16.0084 10 6 )(155 106 )
0.100
24.8 103 N m
My
24.8 kN m
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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450
PROBLEM 4.5
0.1 in.
0.5 in.
M1
Using an allowable stress of 16 ksi, determine the largest couple that can be
applied to each pipe.
(a)
0.2 in.
0.5 in.
M2
(b)
SOLUTION
(a)
I
c
ro4 ri4
4
0.6 in.
Mc
:
I
M
4
(0.64
I
c
0.54 )
52.7 10
3
in 4
(16)(52.7 10 3 )
0.6
M
(b)
I
c
(0.74
4
0.7 in.
Mc
:
I
0.54 )
M
139.49 10
I
c
3
1.405 kip in.
in 4
(16)(139.49 10 3 )
0.7
M
3.19 kip in.
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451
PROBLEM 4.6
r 5 20 mm
A
30 mm
B
M = 2.8 kN · m
Knowing that the couple shown acts in a vertical plane,
determine the stress at (a) point A, (b) point B.
30 mm
120 mm
SOLUTION
I
1
(0.120 m)(0.06 m)3
12
2.1391 10
(a)
A
6
2
1
12
4
(0.02 m) 4
mm 4
(2.8 103 N m)(0.03 m)
2.1391 10 6 mm 4
M yA
I
39.3 MPa
A
(b)
B
M yB
I
(2.8 103 N m)(0.02 m)
2.1391 10 6 m 4
B
26.2 MPa
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452
PROBLEM 4.7
y
Two W4 13 rolled sections are welded together as shown. Knowing that for the steel
alloy used Y 36 ksi and U 58 ksi and using a factor of safety of 3.0, determine the
largest couple that can be applied when the assembly is bent about the z axis.
z
C
SOLUTION
Properties of W4
13 rolled section.
(See Appendix C.)
Area
3.83 in 2
Width
4.060 in.
Iy
3.86 in 4
For one rolled section, moment of inertia about axis b-b is
For both sections,
Ad 2
Ib
Iy
Iz
2 Ib
c
all
M all
width
U
F .S .
all I
c
3.86 (3.83)(2.030)2
19.643 in 4
39.286 in 4
4.060 in.
58
19.333 ksi
3.0
(19.333)(39.286)
4.060
Mc
I
M all 187.1 kip in.
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453
PROBLEM 4.8
y
C
Two W4 13 rolled sections are welded together as shown. Knowing that for the steel
alloy used U 58 ksi and using a factor of safety of 3.0, determine the largest couple
that can be applied when the assembly is bent about the z axis.
z
SOLUTION
Properties of W4
13 rolled section.
(See Appendix C.)
Area
3.83 in 2
Depth
4.16 in.
Ix
11.3 in 4
For one rolled section, moment of inertia about axis a-a is
For both sections,
Ad 2
Ia
Ix
Iz
2Ia
c
all
M all
depth
U
F .S .
all I
c
11.3 (3.83)(2.08)2
27.87 in 4
55.74 in 4
4.16 in.
58
19.333 ksi
3.0
(19.333)(55.74)
4.16
Mc
I
M all
259 kip in.
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454
PROBLEM 4.9
3 in. 3 in. 3 in.
6 in.
Two vertical forces are applied to a beam of the cross section shown. Determine
the maximum tensile and compressive stresses in portion BC of the beam.
2 in.
A
15 kips
15 kips
B
C
40 in.
60 in.
D
40 in.
SOLUTION
A
y0
A y0
18
5
90
18
1
18
36
Y0
108
36
108
3 in.
Neutral axis lies 3 in. above the base.
I1
I2
I
ytop
M
M
1
1
b1h13 A1d12
(3)(6)3 (18)(2) 2 126 in 4
12
12
1
1
3
2
b2 h2 A2 d 2
(9)(2)3 (18)(2) 2 78 in 4
12
12
I1 I 2 126 78 204 in 4
5 in. ybot
Pa 0
Pa (15)(40) 600 kip in.
M ytop
top
bot
3 in.
I
(600)(5)
204
M ybot
I
(600)( 3)
204
top
14.71 ksi (compression)
bot
8.82 ksi (tension)
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455
PROBLEM 4.10
8 in.
1 in.
Two vertical forces are applied to a beam of the cross section shown.
Determine the maximum tensile and compressive stresses in portion
BC of the beam.
6 in.
1 in.
1 in.
4 in.
A
25 kips
25 kips
B
C
20 in.
60 in.
D
20 in.
SOLUTION
A
y0
A y0
8
7.5
60
6
4
24
4
0.5
18
86
86
18
4.778 in.
(8)(2.772) 2
59.94 in 4
(6)(0.778) 2
21.63 in 4
(4)(4.278) 2
73.54 in 4
Yo
2
Neutral axis lies 4.778 in. above the base.
I1
I2
I3
I
ytop
1
1
(8)(1)3
b1h13 A1d12
12
12
1
1
(1)(6)3
b2 h23 A2 d 22
12
12
1
1
(4)(1)3
b3 h33 A3 d32
12
12
I1 I 2 I 3 59.94 21.63
4.778 in.
3.222 in. ybot
M
M
Pa 0
Pa (25)(20) 500 kip in.
Mytop
top
bot
73.57 155.16 in 4
I
Mybot
I
(500)(3.222)
155.16
(500)( 4.778)
155.16
top
10.38 ksi (compression)
bot
15.40 ksi (tension)
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456
10 mm
PROBLEM 4.11
10 mm
10 kN
10 kN
B
50 mm
C
A
D
Two vertical forces are applied to a beam of
the cross section shown. Determine the
maximum tensile and compressive stresses
in portion BC of the beam.
10 mm
50 mm
250 mm
150 mm
150 mm
SOLUTION
A, mm 2
y0 , mm
A y0 , mm3
600
30
18 103
600
30
18 103
300
5
1.5 103
37.5 103
1500
Y0
37.5 103
1500
25mm
Neutral axis lies 25 mm above the base.
I1
I3
I
ytop
35 mm 0.035 m
a 150 mm
M
bot
ybot
25 mm
0.025 m
0.150 m P 10 103 N
(10 103 )(0.150) 1.5 103 N m
Pa
Mytop
top
1
(10)(60)3 (600)(5)2 195 103 mm 4 I 2 I1 195 mm 4
12
1
(30)(10)3 (300)(20) 2 122.5 103 mm 4
12
I1 I 2 I 3 512.5 103 mm 4 512.5 10 9 m 4
I
M ybot
I
(1.5 103 )(0.035)
512.5 10 9
102.4 106 Pa
(1.5 103 )( 0.025)
512.5 10 9
73.2 106 Pa
top
102.4 MPa (compression)
bot
73.2 MPa (tension)
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457
PROBLEM 4.12
216 mm
y
z
Knowing that a beam of the cross section shown is bent about a horizontal
axis and that the bending moment is 6 kN m, determine the total force
acting on the shaded portion of the web.
36 mm
54 mm
C
108 mm
72 mm
SOLUTION
The stress distribution over the entire cross section is given by the bending stress formula:
My
I
x
where y is a coordinate with its origin on the neutral axis and I is the moment of inertia of the entire cross
sectional area. The force on the shaded portion is calculated from this stress distribution. Over an area
element dA, the force is
dF
My
dA
I
x dA
The total force on the shaded area is then
F
My
dA
I
dF
M
I
y dA
M * *
y A
I
where y * is the centroidal coordinate of the shaded portion and A* is its area.
d1
54 18 36 mm
d2
54 36 54 36 mm
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458
PROBLEM 4.12 (Continued)
Moment of inertia of entire cross section:
I1
I2
I
1
1
(216)(36)3 (216)(36)(36)2 10.9175 106 mm4
b1h13 A1d12
12
12
1
1
b2 h23 A2 d 22
(72)(108)3 (72)(108)(36)2 17.6360 106 mm 4
12
12
I1 I 2 28.5535 106 mm 4 28.5535 10 6 m 4
For the shaded area,
A*
(72)(90)
y*
45 mm
A* y *
F
6480 mm 2
291.6 103 mm3
MA* y *
I
291.6 10 6 m
(6 103 )(291.6 10 6 )
28.5535 10 6
61.3 103 N
F
61.3 kN
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459
20 mm
12 mm
PROBLEM 4.13
20 mm
y
12 mm
24 mm
z
Knowing that a beam of the cross section shown is bent about a horizontal axis
and that the bending moment is 4 kN m, determine the total force acting on
the shaded portion of the beam.
20 mm
C
20 mm
24 mm
SOLUTION
Dimensions in mm:
Iz
1
1
(12 12)(88)3
(40)(40)3
12
12
1.3629 106 0.213 106
1.5763 106 mm 4
1.5763 10 6 m 4
For use in Prob. 4.14,
Iy
1
1
(88)(64)3
(24 24)(40)3
12
12
1.9224 106 0.256 106
1.6664 106 mm 4
Bending about horizontal axis.
Mz
1.6664 10 6 m 4
4 kN m
A
M zc
Iz
(4 kN m)(0.044 m)
1.5763 10 6 m 4
111.654 MPa
B
M zc
Iz
(4 kN m)(0.020 m)
1.5763 10 6 m 4
50.752 MPa
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460
PROBLEM 4.13 (Continued)
A (44)(12) 528 mm 2
Portion (1):
avg
1
2
Force1
A
avg A
avg
Force2
Total force on shaded area
1
2
B
avg A
6
m2
1
(111.654) 55.83 MPa
2
(55.83 MPa)(528 10
A (20)(20)
Portion (2):
528 10
400 mm 2
1
(50.752)
2
6
400 10
m2 )
6
29.477 kN
m2
25.376 MPa
(25.376 MPa)(400 10
6
m 2 ) 10.150 kN
29.477 10.150 39.6 kN
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461
20 mm
12 mm
PROBLEM 4.14
20 mm
y
12 mm
Solve Prob. 4.13, assuming that the beam is bent about a vertical axis by a
couple of moment 4 kN m.
24 mm
z
PROBLEM 4.13. Knowing that a beam of the cross section shown is bent
about a horizontal axis and that the bending moment is 4 kN m, determine the
total force acting on the shaded portion of the beam.
20 mm
C
20 mm
24 mm
SOLUTION
My
Bending about vertical axis.
Iy
See Prob. 4.13 for sketch and
4 kN m
1.6664 10
Mc
Iy
(4 kN m)(0.032 m)
1.6664 10 6 m 4
76.81 MPa
E
Mc
Iy
(4 kN m)(0.020 m)
1.6664 10 6 m 4
48.01 MPa
avg
Force1
1
(
2
E)
D
avg A
avg
Force2
Total force on shaded area
1
2
E
avg A
528 10
6
1
(76.81 48.01)
2
(62.41 MPa)(528 10
A (20)(20)
Portion (2):
m4
D
A (44)(12) 528 mm 2
Portion (1):
6
400 mm 2
1
(48.01)
2
6
400 10
62.41 MPa
m 2 ) 32.952 kN
6
m2
6
m 2 ) 9.602 kN
24.005 MPa
(24.005 MPa)(400 10
32.952 9.602
m2
42.6 kN
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462
0.5 in.
0.5 in.
0.5 in.
1.5 in.
1.5 in.
PROBLEM 4.15
Knowing that for the extruded beam shown the allowable stress is 12 ksi
in tension and 16 ksi in compression, determine the largest couple M that
can be applied.
1.5 in.
0.5 in.
M
SOLUTION
A
y0
Ay0
2.25
1.25
2.8125
2.25
0.25
0.5625
4.50
Y
3.375
3.375
4.50
0.75 in.
The neutral axis lies 0.75 in. above bottom.
ytop
I1
I2
I
2.0 0.75 1.25 in.,
ybot
0.75 in.
1
1
b1h13 A1d12
(1.5)(1.5)3 (2.25)(0.5)2 0.984375 in 4
12
12
1
1
2
2
b2 h2 A2 d 2
(4.5)(0.5)3 (2.25)(0.5)2 0.609375 in 4
12
12
I1 I 2 1.59375 in 4
My
I
M
I
y
Top: (compression)
M
(16)(1.59375)
1.25
20.4 kip in.
Bottom: (tension)
M
(12)(1.59375)
0.75
25.5 kip in.
M all
Choose the smaller as Mall.
20.4 kip in.
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463
PROBLEM 4.16
40 mm
The beam shown is made of a nylon for which the allowable stress is
24 MPa in tension and 30 MPa in compression. Determine the largest
couple M that can be applied to the beam.
15 mm
d ! 30 mm
20 mm
M
SOLUTION
Σ
Y0
A, mm2
y0 , mm
A y0 , mm3
600
22.5
13.5 103
300
7.5
2.25 103
15.75 103
900
15.5 103
900
17.5 mm
ytop
ybot
I1
I2
I
| |
The neutral axis lies 17.5 mm above the bottom.
30 17.5 12.5 mm
17.5 mm
0.0175 m
1
b1h13 A1d12
12
1
b2 h23 A2 d 22
12
I1 I 2 61.875
My
I
M
0.0125 m
1
(40)(15)3 (600)(5)2 26.25 103 mm 4
12
1
(20)(15)3 (300)(10)2 35.625 103 mm 4
12
103 mm 4 61.875 10 9 m 4
I
y
Top: (tension side)
M
(24 106 )(61.875 10 9 )
0.0125
Bottom: (compression)
M
(30 106 )(61.875 10 9 )
106.1 N m
0.0175
Choose smaller value.
118.8 N m
M
106.1 N m
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464
PROBLEM 4.17
40 mm
Solve Prob. 4.16, assuming that d
15 mm
d ! 30 mm
40 mm.
PROBLEM 4.16 The beam shown is made of a nylon for which the
allowable stress is 24 MPa in tension and 30 MPa in compression.
Determine the largest couple M that can be applied to the beam.
20 mm
M
SOLUTION
A, mm2
y0 , mm
A y0 , mm3
600
32.5
19.5 103
500
12.5
6.25 103
Σ
Y0
25.75 103
1100
25.75 103
1100
ytop
ybot
I1
I2
I
| |
23.41 mm
The neutral axis lies 23.41 mm above the bottom.
40 23.41 16.59 mm
23.41 mm
0.01659 m
0.02341 m
1
1
b1h13 A1d12
(40)(15)3 (600)(9.09) 2 60.827 103 mm 4
12
12
1
1
b2 h22 A2 d 22
(20)(25)3 (500)(10.91) 2 85.556 103 mm 4
12
12
I1 I 2 146.383 103 mm 4 146.383 10 9 m 4
My
I
M
I
y
Top: (tension side)
M
(24 106 )(146.383 10 9 )
0.01659
212 N m
Bottom: (compression)
M
(30 106 )(146.383 10 9 )
0.02341
187.6 N m
Choose smaller value.
M
187.6 N m
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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465
PROBLEM 4.18
2.4 in.
0.75 in.
Knowing that for the beam shown the allowable stress is 12 ksi in tension and
16 ksi in compression, determine the largest couple M that can be applied.
1.2 in.
M
SOLUTION
rectangle
semi-circular cutout
A1
2.88 in 2
(2.4)(1.2)
A2
2
(0.75)2
A
2.88
y1
0.6 in.
0.8836 in 2
1.9964 in 2
0.8836
4r
(4)(0.75)
0.3183 in.
3
3
Ay
(2.88)(0.6) (0.8836)(0.3183)
A
1.9964
y2
Y
0.7247 in.
Neutral axis lies 0.7247 in. above the bottom.
Moment of inertia about the base:
1 3
bh
3
Ib
8
r4
1
(2.4)(1.2)3
3
(0.75) 4
1.25815 in 4
1.25815 (1.9964)(0.7247) 2
0.2097 in 4
8
Centroidal moment of inertia:
I
Ib
AY 2
ytop
1.2 0.7247
ybot
0.7247 in.
| |
My
I
M
0.4753 in.,
I
y
Top: (tension side)
M
(12)(0.2097)
0.4753
5.29 kip in.
Bottom: (compression)
M
(16)(0.2097)
0.7247
4.63 kip in.
M
Choose the smaller value.
4.63 kip in.
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466
PROBLEM 4.19
80 mm
Knowing that for the extruded beam shown the allowable stress is
120 MPa in tension and 150 MPa in compression, determine the largest
couple M that can be applied.
54 mm
40 mm
M
SOLUTION
Σ
Y
A, mm2
y0 , mm
2160
27
58,320
3
1080
36
38,880
3
A y0 , mm3
3240
d, mm
97,200
97, 200
3240
30 mm
The neutral axis lies 30 mm above the bottom.
ytop
54 30
24 mm
I1
I2
I
| |
1
b1h13 A1d12
12
1
b2 h22 A2 d 22
36
I1 I 2 758.16
My
I
0.024 m
ybot
30 mm
0.030 m
1
(40)(54)3 (40)(54)(3) 2 544.32 103 mm 4
12
1
1
(40)(54)3
(40)(54)(6)2 213.84 103 mm 4
36
2
103 mm 4 758.16 10 9 m 4
I
y
|M |
Top: (tension side)
M
(120 106 )(758.16 10 9 )
0.024
3.7908 103 N m
Bottom: (compression)
M
(150 106 )(758.16 10 9 )
0.030
3.7908 103 N m
Choose the smaller as Mall.
M all
3.7908 103 N m
M all
3.79 kN m
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467
PROBLEM 4.20
48 mm
Knowing that for the extruded beam shown the allowable stress is
120 MPa in tension and 150 MPa in compression, determine the largest
couple M that can be applied.
48 mm
36 mm
48 mm
36 mm
M
SOLUTION
A, mm2
y0 , mm
Solid rectangle
4608
48
221,184
Square cutout
–1296
30
–38,880
Σ
Y
3312
182,304
3312
ytop
ybot
I1
I2
I
| |
55.04 mm
182,304
Neutral axis lies 55.04 mm above bottom.
96 55.04 40.96 mm
55.04 mm
0.04096 m
0.05504 m
1
b1h13 A1d12
12
1
b2 h32 A2 d 22
12
I1 I 2 2.8147
My
I
A y0 , mm3
M
1
(48)(96)3 (48)(96)(7.04)2 3.7673 106 mm 4
12
1
(36)(36)3 (36)(36)(25.04) 2 0.9526 106 mm 4
12
106 mm 4 2.8147 10 6 m 4
I
y
Top: (tension side)
M
(120 106 )(2.8147 10 6 )
0.04096
Bottom: (compression)
M
(150 106 )(2.8147 106 )
0.05504
Mall is the smaller value.
M
7.67 103 N m
8.25 103 N m
7.67 103 N m
7.67 kN m
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468
PROBLEM 4.21
Straight rods of 6-mm diameter and 30-m length are stored by coiling the
rods inside a drum of 1.25-m inside diameter. Assuming that the yield
strength is not exceeded, determine (a) the maximum stress in a coiled rod,
(b) the corresponding bending moment in the rod. Use E 200 GPa.
SOLUTION
Let
D
inside diameter of the drum,
d
diameter of rod,
1
d,
2
radius of curvature of center line of rods when bent.
1
D
2
I
(a)
(b)
Ec
max
M
EI
4
c4
c
1
d
2
4
(200 109 )(0.003)
0.622
(200 109 )(63.617 10
0.622
1
(1.25)
2
(0.003) 4
1
(6 10 3 )
2
63.617 10
12
0.622 m
m4
965 106 Pa
12
)
965 MPa
20.5 N m
M
20.5 N m
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469
M'
PROBLEM 4.22
M
8 mm
A 900-mm strip of steel is bent into a full circle by two couples applied
as shown. Determine (a) the maximum thickness t of the strip if the
allowable stress of the steel is 420 MPa, (b) the corresponding moment
M of the couples. Use E 200 GPa.
t
r
900 mm
SOLUTION
When the rod is bent into a full circle, the circumference is 900 mm. Since the circumference is equal to
2 times , the radius of curvature, we get
900 mm
2
Stress:
E
Ec
420 MPa and E
For
c
(a)
143.24 mm
0.14324 m
or
c
E
200 GPa,
(0.14324)(420 106 )
200 109
t
Maximum thickness:
0.3008 10
2c
3
m
0.6016 10
3
m
t
0.602 mm
Moment of inertia for a rectangular section.
I
(b)
bt 3
12
Bending moment:
M
(8 10 3 )(0.6016 10 3 )3
12
M
145.16 10
15
m4
EI
(200 109 )(145.16 10
0.14324
15
)
0.203 N m
M
0.203 N m
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470
PROBLEM 4.23
Straight rods of 0.30-in. diameter and 200-ft length are sometimes
used to clear underground conduits of obstructions or to thread wires
through a new conduit. The rods are made of high-strength steel and,
for storage and transportation, are wrapped on spools of 5-ft
diameter. Assuming that the yield strength is not exceeded,
determine (a) the maximum stress in a rod, when the rod, which is
initially straight, is wrapped on a spool, (b) the corresponding
bending moment in the rod. Use E 29 106 psi .
5 ft
SOLUTION
Radius of cross section:
r
Moment of inertia:
I
(a)
(b)
D
5 ft
c
r
60 in.
4
r4
1
(0.30)
2
4
(0.15) 4
1
D
2
30 in.
106 )(0.15)
30
145.0
0.15 in.
397.61 10
6
in 4
0.15 in.
Ec
(29
max
M
1
d
2
EI
103 psi
(29 106 )(397.61 10 6 )
30
max
M
145.0 ksi
384 lb in.
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471
PROBLEM 4.24
12 mm
y
60 N · m
A 60-N m couple is applied to the steel bar shown. (a) Assuming that
the couple is applied about the z axis as shown, determine the maximum
stress and the radius of curvature of the bar. (b) Solve part a, assuming
that the couple is applied about the y axis. Use E 200 GPa.
20 mm
z
SOLUTION
(a)
Bending about z-axis.
I
c
1
(b)
1 3 1
bh
(12)(20)3 8 103 mm 4
12
12
20
10 mm 0.010 m
2
Mc
I
(60)(0.010)
8 10 9
M
EI
60
(200 109 )(8 10 9 )
8 10 9 m 4
75.0 106 Pa
37.5 10 3 m
75.0 MPa
1
26.7 m
Bending about y-axis.
I
c
1
1 3 1
bh
(20)(12)3 2.88 103 mm 4
12
12
12
6 mm 0.006 m
2
Mc (60)(0.006)
125.0 106 Pa
9
I
2.88 10
M
EI
60
(200 10 )(2.88 10 9 )
9
2.88 10 9 m 4
104.17 10 3 m
125.0 MPa
1
9.60 m
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472
10 mm
M
(a) Using an allowable stress of 120 MPa, determine the largest couple
M that can be applied to a beam of the cross section shown. (b) Solve
part a, assuming that the cross section of the beam is an 80-mm square.
80 mm
C
PROBLEM 4.25
10 mm
5 mm
80 mm
5 mm
SOLUTION
(a)
I I1 4 I 2 , where I1 is the moment of inertia of an 80-mm square and I2 is the moment of inertia of
one of the four protruding ears.
I1
I2
I
1 3
bh
12
1
(80)(80)3
12
3.4133 106 mm 4
1 3
1
(5)(10)3 (5)(10)(45)2 101.667 103 mm 4
bh Ad 2
12
12
I1 4 I 2 3.82 106 mm 4 3.82 10 6 mm 4 ,
c 50 mm 0.050 m
Mc
I
I
c
M
(120 106 )(3.82 10 6 )
0.050
9.168 103 N m
9.17 kN m
(b)
Without the ears:
I
M
I1
I
c
3.4133 10
6
m2 ,
c
40 mm
0.040 m
(120 106 )(3.4133 10 6 )
10.24 103 N m
0.040
10.24 kN m
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473
0.1 in.
PROBLEM 4.26
0.2 in.
A thick-walled pipe is bent about a horizontal axis by a couple M. The
pipe may be designed with or without four fins. (a) Using an allowable
stress of 20 ksi, determine the largest couple that may be applied if the
pipe is designed with four fins as shown. (b) Solve part a, assuming that
the pipe is designed with no fins.
1.5 in.
M
0.75 in.
SOLUTION
I x of hollow pipe:
Ix
I x of fins:
Ix
(1.5 in.) 4
4
2
(0.75 in.) 4
1
(0.1)(0.2)3
12
3.7276 in 4
(0.1 0.2)(1.6) 2
2
1
(0.2)(0.1)3
12
0.1026 in 4
(a)
(b)
Pipe as designed, with fins:
Ix
3.8302 in 4 ,
all
20 ksi,
c 1.7 in.
M
all
Ix
c
(20 ksi)
3.8302 in 4
1.7 in.
M
45.1 kip in.
M
49.7 kip in.
Pipe with no fins:
all
M
20 ksi,
all
Ix
c
Ix
3.7276 in 4 ,
(20 ksi)
c 1.5 in.
3.7276 in 4
1.5 in.
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474
PROBLEM 4.27
M
A couple M will be applied to a beam of rectangular cross section that
is to be sawed from a log of circular cross section. Determine the ratio
d/b for which (a) the maximum stress m will be as small as possible,
(b) the radius of curvature of the beam will be maximum.
M'
d
b
SOLUTION
Let D be the diameter of the log.
D2
m
d2
1
bd 3
12
I
(a)
b2
c
is the minimum when
I
c
d
db
D2
1
d
2
I
c
b2
D2
1 2
bd
6
I
is maximum.
c
1
1 2
b( D 2 b 2 )
Db
6
6
I
1 2 3 2
0
D
b
6
6
c
d
(b)
d2
1 2
D
3
1 3
b
6
1
D
3
b
2
D
3
d
b
2
d
b
3
EI
M
is maximum when I is maximum,
(D2
d 2 )d 6 is maximum.
6 D 2d 5
b
1
bd 3 is maximum, or b2d 6 is maximum.
12
D2
8d 7
0
3 2
D
4
d
1
D
2
3
D
2
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475
PROBLEM 4.28
h0
M
h
C
h0
h
A portion of a square bar is removed by milling, so that its cross section
is as shown. The bar is then bent about its horizontal axis by a couple M.
Considering the case where h 0.9h0, express the maximum stress in the
bar in the form m k 0 , where 0 is the maximum stress that would
have occurred if the original square bar had been bent by the same
couple M, and determine the value of k.
SOLUTION
I
4 I1
2I2
1
h h3
12
1 4 4
h
h 0 h3
3
3
h
Mc
Mh
4 h h3
I
3 0
(4)
c
For the original square,
h
h0 , c
0
3M
(4h0 3h0 )h02
1
(2h0 2h)(h3 )
3
4
4
h h3
h 0 h3 h 4
3
3
(2)
h
4
3M
3h) h 2
h0 .
3M
h03
h03
0
(4h 0
(4h0
0.950
h03
3h)h 2
(4h0
(3)(0.9)h0 )(0.9 h02 )
0.950
k
0
0.950
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476
PROBLEM 4.29
h0
M
In Prob. 4.28, determine (a) the value of h for which the maximum stress
m is as small as possible, (b) the corresponding value of k.
h
PROBLEM 4.28 A portion of a square bar is removed by milling, so that
its cross section is as shown. The bar is then bent about its horizontal
axis by a couple M. Considering the case where h 0.9h0, express the
maximum stress in the bar in the form m k 0 , where 0 is the
maximum stress that would have occurred if the original square bar had
been bent by the same couple M, and determine the value of k.
C
h
h0
SOLUTION
I
4 I1
2I 2
1
1
hh3 (2)
(2h0 2h) h3
12
3
1 4 4
4 3 4
h
h 0 h3
h
h 0 h3 h 4
3
3
3
3
I 4
2
3
h
h0 h h
c 3
(4)
c
d 4
I
is maximum at
h0 h2
c
dh 3
h3
0.
8
h 0 h 3h 2
3
I
c
For the original square,
4
8
h0
h0
3
9
h
h0
2
8
h0
9
3
256 3
h0
729
c
h0
I0
c0
0
Mc0
I0
3M
h02
0
0
729 1
256 3
h
Mc
I
8
h0
9
729M
256h03
1 3
h0
3
729
768
0.949
k
0.949
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477
PROBLEM 4.30
For the bar and loading of Concept Application 4.1, determine (a) the radius of curvature , (b) the radius of
of a transverse cross section, (c) the angle between the sides of the bar that were originally
curvature
vertical. Use E 29 106 psi and v 0.29.
SOLUTION
M
From Example 4.01,
(a)
(b)
1
1
1
(c)
M
EI
vc
v
v
(30 103 )
(29 106 )(1.042)
1
v
993 10
6
in.
1.042 in 4
I
1
1007 in.
c
(0.29)(993 10 6 )in.
length of arc
radius
30 kip in.
b
0.8
3470
1
288 10
6
in.
230 10 6 rad
1
3470 in.
0.01320
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478
PROBLEM 4.31
y
A
z
M
A W200 31.3 rolled-steel beam is subjected to a couple M of moment
45 kN m. Knowing that E 200 GPa and v 0.29, determine (a) the
of a transverse cross
radius of curvature , (b) the radius of curvature
section.
C
x
SOLUTION
For W 200 31.3 rolled steel section,
I
31.3 106 mm 4
31.3 10 6 m 4
(a)
(b)
1
1
M
EI
v
1
45 103
(200 109 )(31.3 10 6 )
(0.29)(7.1885 10 3 )
7.1885 10 3 m
2.0847 10 3 m
1
1
139.1 m
480 m
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479
PROBLEM 4.32
y
!
2
y # %c
"y
It was assumed in Sec. 4.1B that the normal stresses y in a
member in pure bending are negligible. For an initially straight
elastic member of rectangular cross section, (a) derive an
approximate expression for y as a function of y, (b) show that
(c /2 )( x ) max and, thus, that y can be neglected in
( y )max
all practical situations. (Hint: Consider the free-body diagram of
the portion of beam located below the surface of ordinate y and
assume that the distribution of the stress x is still linear.)
!
2
"y
"x
"x
!
2
!
2
y # $c
SOLUTION
Denote the width of the beam by b and the length by L.
L
cos
Using the free body diagram above, with
Fy
0:
2
sin
L
2
y
But,
(a)
(
y
x ) max
y
c
c
y dy
The maximum value
y bL
x
(
x ) max
(
x ) max
y2
2
c
y
occurs at y
y
2
c
y
c
2
x
1
x b dy
sin
0
2
y
dy
L
c
x
1
dy
y
c
x
dy
y
c
y
(
y
c
x ) max
2 c
( y2
c2 )
0.
(b)
(
y ) max
(
x ) max c
2 c
2
(
x ) max c
2
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480
PROBLEM 4.33
Brass
6 mm
A bar having the cross section shown has been formed by securely
bonding brass and aluminum stock. Using the data given below,
determine the largest permissible bending moment when the
composite bar is bent about a horizontal axis.
Aluminum
30 mm
6 mm
Aluminum
Brass
70 GPa
105 GPa
100 MPa
160 MPa
Modulus of elasticity
30 mm
Allowable stress
SOLUTION
Use aluminum as the reference material.
n
1.0 in aluminum
n
Eb /Ea
105/70
1.5 in brass
For the transformed section,
I1
n1
b1h13 n1 A1d 12
12
1.5
(30)(6)3 (1.5)(30)(6)(18)3
12
I2
n2
b2h23
12
I3
I1
88.29 103 mm 4
I
I1
I2
1.0
(30)(30)3
12
I3
Aluminum:
n 1.0,
M
Brass:
Choose the smaller value
67.5 103 mm 4
244.08 103 mm 4
244.08 10
nMy
I
M
y 15 mm
9
y
21 mm
m4
I
ny
0.015 m,
(100 106 )(244.08 10 9 )
(1.0)(0.015)
n 1.5,
88.29 103 mm 4
0.021 m,
M
(160 106 )(244.08 10 9 )
(1.5)(0.021)
M
1.240 103 N m
100 106 Pa
1.627 103 N m
160 106 Pa
1.240 103 N m
1.240 kN m
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481
8 mm
PROBLEM 4.34
8 mm
32 mm
8 mm
32 mm
A bar having the cross section shown has been formed by securely
bonding brass and aluminum stock. Using the data given below,
determine the largest permissible bending moment when the
composite bar is bent about a horizontal axis.
8 mm
Brass
Aluminum
Brass
70 GPa
105 GPa
100 MPa
160 MPa
Modulus of elasticity
Aluminum
Allowable stress
SOLUTION
Use aluminum as the reference material.
For aluminum, n 1.0
For brass, n
Eb /Ea
105/70 1.5
Values of n are shown on the sketch.
For the transformed section,
I1
I2
I3
I
| |
Aluminum:
I1
I2
nMy
I
I 3 141.995 103 mm 4
M
100 106 Pa
0.016 m,
(100 106 )(141.995 10 9 )
(1.0)(0.016)
n 1.5, | y | 16 mm
M
141.995 10 9 m 4
I
ny
n 1.0, | y | 16 mm
M
Brass:
n1
1.5
b1h13
(8)(32)3 32.768 103 mm 4
12
12
n2
1.0
b2 H 23 h23
(32)(323 163 ) 76.459 103 mm 4
12
12
I1 32.768 103 mm 4
887.47 N m
160 106 Pa
0.016 m,
(160 106 )(141.995 10 9 )
(1.5)(0.016)
946.63 N m
Choose the smaller value.
M
887 N m
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482
PROBLEM 4.35
Brass
6 mm
For the composite bar indicated, determine the largest permissible
bending moment when the bar is bent about a vertical axis.
30 mm
PROBLEM 4.35. Bar of Prob. 4.33.
Aluminum
Modulus of elasticity
6 mm
Allowable stress
30 mm
Aluminum
Brass
70 GPa
105 GPa
100 MPa
160 MPa
SOLUTION
Use aluminum as reference material.
n
1.0 in aluminum
n
Eb /Ea
105/70
1.5 in brass
For transformed section,
I1
I2
n1
b1h13
12
1.5
(6)(30)3
12
n2
b2h23
12
1.0
(30)(30)3
12
I1
20.25 103 mm 4
I
I1
I2
n 1.0,
M
Brass:
67.5 103 mm 4
I3
nMy
I
Aluminum:
20.25 103 mm 4
108
103 mm 4
108
10
9
m4
I
ny
M
y 15 mm
0.015 m,
(100 106 )(108 10 9 )
(1.0)(0.015)
n 1.5,
M
I3
y 15 mm
720 N m
0.015 m,
(160 106 )(108 10 9 )
(1.5)(0.015)
100 106 Pa
160 106 Pa
768 N m
M
Choose the smaller value.
720 N m
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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483
8 mm
PROBLEM 4.36
8 mm
32 mm
8 mm
For the composite bar indicated, determine the largest permissible
bending moment when the bar is bent about a vertical axis.
PROBLEM 4.36 Bar of Prob. 4.34.
32 mm
8 mm
Brass
Modulus of elasticity
Allowable stress
Aluminum
Aluminum
70 GPa
100 MPa
Brass
105 GPa
160 MPa
SOLUTION
Use aluminum as the reference material.
For aluminum, n 1.0
For brass, n
Eb /Ea
105/70 1.5
Values of n are shown on the sketch.
For the transformed section,
I1
I2
I3
I
| |
Aluminum:
I1
I2
nMy
I
I3
M
354.99 103 mm 4
354.99 10 9 m 4
I
ny
n 1.0, | y | 16 mm
M
Brass:
n1
1.5
h1 B13 b13
(32)(483 323 ) 311.296 103 mm 4
12
12
n2
1.0
h2b23
(8)(32)3 21.8453 103 mm 4
12
12
I 2 21.8453 103 mm 4
0.016 m,
(100 106 )(354.99 10 9 )
(1.0)(0.016)
n 1.5 | y | 24 mm
0.024 m
M
(160 106 )(354.99 10 9 )
(1.5)(0.024)
M
1.57773 103 N m
100 106 Pa
2.2187 103 N m
160 106 Pa
1.57773 103 N m
Choose the smaller value.
M
1.578 kN m
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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484
PROBLEM 4.37
1
5 3 2 in.
Wooden beams and steel plates are securely bolted together to form the composite
member shown. Using the data given below, determine the largest permissible bending
moment when the member is bent about a horizontal axis.
10 in.
Modulus of elasticity:
1
5 3 2 in.
6 in.
Allowable stress:
Wood
Steel
2 106 psi
29 106 psi
2000 psi
22 ksi
SOLUTION
Use wood as the reference material.
n 1.0 in wood
n
E s /E w
29/2 14.5 in steel
For the transformed section,
I1
n1
b1h13
12
n1 A1d12
3
I2
I3
I
nMy
I
Wood:
n
M
Steel:
n
M
M
I
ny
1.0,
y
14.5
1
1
(5)
(14.5)(5)
(5.25)2
12
2
2
n2
1.0
b2 h22
(6)(10)3 500 in 4
12
12
I1 999.36 in 4
I1
y
2498.7 in 4
2000 psi
999.5 103 lb in.
5.5 in.,
(22 103 )(2499)
(14.5)(5.5)
Choose the smaller value.
I3
5 in.,
(2000)(2499)
(1.0)(5)
14.5,
I2
999.36 in 4
22 ksi
22 103 psi
689.3 103 lb in.
M
689 103 lb in.
M
689 kip in.
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
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485
PROBLEM 4.38
10 in.
3 in.
1
2
Wooden beams and steel plates are securely bolted together to form the composite
member shown. Using the data given below, determine the largest permissible bending
moment when the member is bent about a horizontal axis.
3 in.
Modulus of elasticity:
in.
Allowable stress:
Wood
Steel
2 106 psi
29 106 psi
2000 psi
22 ksi
SOLUTION
Use wood as the reference material.
n 1.0 in wood
n
E s /E w
29/2 14.5 in steel
For the transformed section,
I1
I2
I3
I
nMy
I
Wood:
n
M
Steel:
n
M
M
I
ny
1.0,
y
n1
1.0
(3)(10)3 250 in 4
b1h13
12
12
n2
14.5 1
b2 h23
(10)3 604.17 in 4
12
12 2
I1
y
I3
1104.2 in 4
2000 psi
441.7 103 lb in.
5 in.,
(22 103 )(1104.2)
(14.5)(5)
Choose the smaller value.
I2
5 in.,
(2000)(1104.2)
(1.0)(5)
14.5,
250 in 4
I1
22 ksi
22 103 psi
335.1 103 lb in.
M
335 103 lb in.
M
335 kip in.
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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486
PROBLEM 4.39
A copper strip (Ec = 105 GPa) and an aluminum strip (Ea = 75 GPa)
are bonded together to form the composite beam shown. Knowing
that the beam is bent about a horizontal axis by a couple of moment
M = 35 N m, determine the maximum stress in (a) the aluminum
strip, (b) the copper strip.
SOLUTION
Use aluminum as the reference material.
n 1.0 in aluminum
n
Ec /Ea
105/75 1.4 in copper
Transformed section:
A, mm2
nA, mm 2
y0 , mm
nAy0 , mm3
144
144
9
1296
144
201.6
3
Σ
Y0
345.6
1900.8
345.6
604.8
1900.8
5.50 mm
The neutral axis lies 5.50 mm above the bottom.
I
n1
1.0
b1h13 n1 A1d12
(24)(6)3 (1.0)(24)(6)(3.5) 2 2196 mm 4
12
12
n2
1.4
(24)(6)3 (1.4)(24)(6)(2.5)2 1864.8 mm 4
b2 h23 n2 A2 d 22
12
12
I1 I 2 4060.8 mm 4 4.0608 10 9 m 4
n
1.0,
I1
I2
(a)
Aluminum:
y
nMy
I
12
5.5
6.5 mm
0.0065 m
(1.0)(35)(0.0065)
4.0608 10 9
56.0 106 Pa
56.0 MPa
56.0 MPa
(b)
Copper:
n
1.4,
y
nMy
I
5.5 mm
0.0055 m
(1.4)(35)( 0.0055)
4.0608 10 9
66.4 106 Pa
66.4 MPa
66.4 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
487
PROBLEM 4.40
A copper strip (Ec = 105 GPa) and an aluminum strip (Ea = 75 GPa)
are bonded together to form the composite beam shown. Knowing
that the beam is bent about a horizontal axis by a couple of moment
M = 35 N m, determine the maximum stress in (a) the aluminum
strip, (b) the copper strip.
SOLUTION
Use aluminum as the reference material.
n 1.0 in aluminum
n
Ec /Ea
105/75 1.4 in copper
Transformed section:
A, mm2
nA, mm 2
Ay0 , mm
nAy0 , mm3
216
216
7.5
1620
72
100.8
1.5
Σ
Y0
151.8
316.8
1771.2
316.8
1771.2
5.5909 mm
The neutral axis lies 5.5909 mm above the bottom.
I
n1
1.0
b1h13 n1 A1d12
(24)(9)3 (1.0)(24)(9)(1.9091)2 2245.2 mm 4
12
12
n2
1.4
b2 h23 n2 A2 d 22
(24)(3)3 (1.4)(24)(3)(4.0909) 2 1762.5 mm 4
12
12
I1 I 2 4839 mm 4 4.008 10 9 m 4
n
1.0,
I1
I2
(a)
Aluminum:
y
nMy
I
12
5.5909
6.4091 mm
(1.0)(35)(0.0064091)
4.008 10 9
0.0064091
56.0 10
6
Pa
56.0 MPa
56.0 MPa
(b)
Copper:
n
1.4,
y
nMy
I
5.5909 mm
0.0055909 m
(1.4)(35)( 0.0055909)
4.008 10 9
68.4 106 Pa
68.4 MPa
68.4 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
488
PROBLEM 4.41
6 in.
M
12 in.
53
1
2
The 6 12-in. timber beam has been strengthened by bolting to it the steel
reinforcement shown. The modulus of elasticity for wood is 1.8 106 psi and for
steel, 29 106 psi. Knowing that the beam is bent about a horizontal axis by a
couple of moment M 450 kip in., determine the maximum stress in (a) the
wood, (b) the steel.
in.
SOLUTION
Use wood as the reference material.
For wood,
n 1
For steel,
n
Es /Ew
wood
Transformed section:
Yo
29 / 1.8 16.1111
421.931
112.278
3.758 in.
steel
A, in 2
nA, in 2
72
72
6
40.278
0.25
2.5
y0
112.278
nA y0 , in 3
432
10.069
421.931
The neutral axis lies 3.758 in. above the wood-steel interface.
I1
I2
I
M
(a)
n1
1
b1h13 n1 A1d12
(6)(12)3 (72)(6 3.758) 2 1225.91 in 4
12
12
n2
16.1111
b2 h23 n2 A2 d 22
(5)(0.5)3 (40.278)(3.578 0.25)2
12
12
I1 I 2 1873.77 in 4
nMy
I
450 kip in.
Wood:
n 1,
y 12 3.758 8.242 in.
w
(b)
Steel:
(1)(450)(8.242)
1873.77
n 16.1111,
s
647.87 in 4
y
1.979 ksi
3.758 0.5
1.979 ksi
w
4.258 in.
(16.1111)(450)( 4.258)
16.48 ksi
1873.77
s
16.48 ksi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
489
PROBLEM 4.42
6 in.
M
12 in.
The 6 12-in. timber beam has been strengthened by bolting to it the steel
reinforcement shown. The modulus of elasticity for wood is 1.8 106 psi and for
steel, 29 106 psi. Knowing that the beam is bent about a horizontal axis by a
couple of moment M 450 kip in., determine the maximum stress in (a) the
wood, (b) the steel.
C8 & 11.5
SOLUTION
Use wood as the reference material.
For wood,
n 1
For steel,
n
Es
Ew
29 106
1.8 106
16.1111
For C8 11.5 channel section,
A 3.38 in 2 , tw
0.220 in., x
0.571 in., I y
1.32 in 4
For the composite section, the centroid of the channel (part 1) lies 0.571 in. above the bottom of the section.
The centroid of the wood (part 2) lies 0.220 6.00 6.22 in. above the bottom.
Transformed section:
A, in2
3.38
Part
1
72
2
nA, in2
54.456
y , in.
0.571
nAy , in 3
31.091
d, in.
3.216
72
6.22
447.84
2.433
478.93
126.456
Y0
478.93 in 3
126.456 in 2
d
3.787 in.
y0
Y0
The neutral axis lies 3.787 in. above the bottom of the section.
I1
I2
I
M
(a)
n1 I1
n1 A1d12
(16.1111)(1.32) (54.456)(3.216) 2
n2
1
b2 h23 n2 A2 d 22
(6)(12)3
12
12
I1 I 2 1874.69 in 4
450 kip in
(72)(2.433) 2
1290.20 in 4
n My
I
y 12 0.220 3.787 8.433 in.
Wood:
n 1,
Steel:
(1)(450)(8.433)
2.02 ksi
1874.69
n 16.1111, y
3.787 in.
w
(b)
s
584.49 in 4
(16.1111)(450)( 3.787)
14.65 ksi
1874.67
w
s
2.02 ksi
14.65 ksi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
490
PROBLEM 4.43
6 mm
Aluminum
Copper
6 mm
For the composite beam indicated, determine the radius of curvature
caused by the couple of moment 35 N m.
Beam of Prob. 4.39.
24 mm
SOLUTION
See solution to Prob. 4.39 for the calculation of I.
1
M
Ea I
35
(75 10 )(4.0608 10 9 )
9
0.1149 m
1
8.70 m
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491
PROBLEM 4.44
Aluminum
9 mm
Copper
3 mm
For the composite beam indicated, determine the radius of curvature
caused by the couple of moment 35 N m.
Beam of Prob. 4.40.
24 mm
SOLUTION
See solution to Prob. 4.40 for the calculation of I.
1
M
Ea I
35
(75 10 )(4.008 10 9 )
9
0.1164 m
1
8.59 m
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
492
6 in.
PROBLEM 4.45
For the composite beam indicated, determine the radius of curvature caused by the
couple of moment 450 kip in.
M
12 in.
Beam of Prob. 4.41.
53
1
2
in.
SOLUTION
See solution to Prob. 4.41 for calculation of I.
I
1873.77 in 4
M
450 kip in
1
M
EI
Ew
1.8 106 psi
450 103 lb in.
450 103
(1.8 106 )(1873.77)
133.421 10 6 in.
1
7495 in. 625 ft
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
493
6 in.
PROBLEM 4.46
M
12 in.
For the composite beam indicated, determine the radius of curvature caused by the
couple of moment 450 kip in.
Beam of Prob. 4.42.
C8 & 11.5
SOLUTION
See solution to Prob. 4.42 for calculation of I.
I
M
1
1874.69 in 4
Ew
1.8 106 psi
450 kip in. 450 103 lb in.
M
EI
450 103
(1.8 106 )(1874.69)
133.355 10 6 in.
1
7499 in. 625 ft
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
494
PROBLEM 4.47
5
8
-in. diameter
4 in.
A concrete slab is reinforced by 85 -in.-diameter steel rods
placed on 5.5-in. centers as shown. The modulus of elasticity is
106 psi for the steel.
3 106 psi for the concrete and 29
Using an allowable stress of 1400 psi for the concrete and
20 ksi for the steel, determine the largest bending moment in a
portion of slab 1 ft wide.
5.5 in.
5.5 in.
5.5 in.
6 in.
5.5 in.
SOLUTION
n
29 106
3 106
Es
Ec
9.6667
Consider a section 5.5 in. wide.
As
4
d s2
5
4 8
2
0.3068 in 2
2.9657 in 2
nAs
Locate the natural axis.
5.5 x
x
2
(4
2.75 x 2
x )(2.9657)
0
2.9657 x 11.8628 0
Solve for x.
x 1.6066 in.
I
M
2.3934 in.
1
(5.5) x3 (2.9657)(4 x)2
3
1
(5.5)(1.6066)3 (2.9657)(2.3934) 2
3
nMy
I
Concrete: n
4 x
1,
y
M
I
ny
1.6066 in.,
(24.591)(1400)
(1.0)(1.6066)
24.591 in 4
1400 psi
21.429
103 lb in.
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
495
PROBLEM 4.47 (Continued)
Steel:
n
M
9.6667,
y
2.3934 in.,
(24.591)(20 103 )
(9.6667)(2.3934)
21.258
20 ksi=20 103 psi
103 lb in.
Choose the smaller value as the allowable moment for a 5.5 in. width.
M
21.258
103 lb in.
For a 1 ft = 12 in. width,
M
12
(21.258
5.5
M
46.38 kip in.
103 )
46.38
103 lb in.
3.87 kip ft
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
496
PROBLEM 4.48
5
8
-in. diameter
Solve Prob. 4.47, assuming that the spacing of the 85 -in.-diameter
steel rods is increased to 7.5 in.
4 in.
PROBLEM 4.47 A concrete slab is reinforced by 85 -in.-diameter
steel rods placed on 5.5-in. centers as shown. The modulus of
elasticity is 3 × 106 psi for the concrete and 29 106 psi for the
steel. Using an allowable stress of 1400 psi for the concrete and
20 ksi for the steel, determine the largest bending moment in a
portion of slab 1 ft wide.
5.5 in.
5.5 in.
5.5 in.
6 in.
5.5 in.
SOLUTION
29 106 psi
3 106 psi
Es
Ec
n
9.667
Number of rails per foot:
12 in.
1.6
7.5 in.
5
5
Area of -in.-diameter bars per foot: 1.6
8
4 8
2
0.4909 in 2
As
Transformed section, all concrete.
First moment of area:
12 x
x
2
4.745(4 x) 0
x 1.4266 in.
nAs
I NA
For concrete:
all
M
For steel:
all
M
We choose the smaller M.
M
1
(12)(1.4266)3
3
1400 psi c
I
c
steel
n
I
c
4.745(4 1.4266)2
4.745 in 2
43.037 in 4
x 1.4266 in.
(1400 psi)
20 ksi c
9.667(0.4909)
4 x
43.037 in 4
1.4266 in.
M
42.24 kip.in.
M
34.60 kip in.
4 1.4266 2.5734 in.
20 ksi 43.042 in 4
9.667 2.5734 in.
34.60 kip in.
M
Steel controls.
2.88 kip ft
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
497
PROBLEM 4.49
540 mm
The reinforced concrete beam shown is subjected to a positive bending
moment of 175 kN m. Knowing that the modulus of elasticity is 25 GPa
for the concrete and 200 GPa for the steel, determine (a) the stress in the
steel, (b) the maximum stress in the concrete.
25-mm
diameter
60 mm
300 mm
SOLUTION
n
As
nAs
Es
Ec
4
200 GPa
25 GPa
4
d2
(4)
8.0
(25) 2
4
1.9635 103 mm 2
15.708 103 mm 2
Locate the neutral axis.
x
(15.708 103 )(480 x) 0
2
15.708 103 x 7.5398 106 0
300 x
150 x 2
Solve for x.
x
15.708 103
x 177.87 mm,
I
(15.708 103 ) 2 (4)(150)(7.5398 106 )
(2)(150)
480 x 302.13 mm
1
(300) x3 (15.708 103 )(480 x)2
3
1
(300)(177.87)3 (15.708 103 )(302.13)2
3
1.9966 109 mm 4 1.9966 10 3 m 4
nMy
I
(a)
Steel:
y
302.45 mm
0.30245 m
(8.0)(175 103 )( 0.30245)
1.9966 10 3
(b)
Concrete:
y 177.87 mm
212 106 Pa
212 MPa
0.17787 m
(1.0)(175 103 )(0.17787)
1.9966 10 3
15.59 106 Pa
15.59 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
498
PROBLEM 4.50
540 mm
Solve Prob. 4.49, assuming that the 300-mm width is increased to 350 mm.
25-mm
diameter
PROBLEM 4.49 The reinforced concrete beam shown is subjected to a
positive bending moment of 175 kN m. Knowing that the modulus of
elasticity is 25 GPa for the concrete and 200 GPa for the steel, determine
(a) the stress in the steel, (b) the maximum stress in the concrete.
60 mm
300 mm
SOLUTION
n
As
Es
Ec
4
4
200 GPa
25 GPa
d2
(4)
4
8.0
(25) 2
1.9635 103 mm 2
nAs
15.708 103 mm 2
Locate the neutral axis.
x
(15.708 103 )(480 x) 0
2
15.708 103 x 7.5398 106 0
350 x
175 x 2
Solve for x.
(15.708 103 ) 2 (4)(175)(7.5398 106 )
(2)(175)
x 167.48 mm, 480 x 312.52 mm
x
I
(a)
Steel:
y
15.708 103
1
(350) x3 (15.708 103 )(480 x)2
3
1
(350)(167.48)3 (15.708 103 )(312.52) 2
3
2.0823 109 mm 4 2.0823 10 3 m 4
nMy
I
312.52 mm
0.31252 m
(8.0)(175 103 )( 0.31252)
2.0823 10 3
(b)
Concrete:
y 167.48 mm
210 106 Pa
210 MPa
0.16748 m
(1.0)(175 103 )(0.16748)
2.0823 10 3
14.08 106 Pa
14.08 MPa
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499
4 in.
24 in.
PROBLEM 4.51
Knowing that the bending moment in the reinforced concrete beam is
100 kip ft and that the modulus of elasticity is 3.625 106 psi for the
concrete and 29 106 psi for the steel, determine (a) the stress in the
steel, (b) the maximum stress in the concrete.
20 in.
1-in.
diameter
2.5 in.
12 in.
SOLUTION
n
Es
Ec
As
(4)
29 106
3.625 106
(1)2
4
8.0
3.1416 in 2
nAs
25.133 in 2
Locate the neutral axis.
96 x
6x2
192
Solve for x.
(24)(4)( x
2)
339.3
25.133x
d3
I1
I2
I3
I
Steel:
17.5
nA3d32
(25.133)(12.350) 2
I1
I3
n
Concrete:
where M
8.0
n
1.0,
c
6x2
4
121.133x
x)
147.3
0
0
1.150 in.
12.350 in.
(24)(4)(3.150) 2
1080.6 in 4
6.1 in 4
3833.3 in 4
4920 in 4
100 kip ft
y
y
1200 kip in.
12.350 in.
(8.0)(1200)( 12.350)
4920
s
(b)
x
1
(12)(1.150)3
3
I2
or
1
(24)(4)3
12
A1d12
(25.133)(17.5
(4)(6)(147.3)
4
1
b1h13
12
1 3
b2 x
3
nMy
I
(a)
0
(121.133)2
(2)(6)
121.133
x
x
2
(12 x)
4
1.150
s
24.1 ksi
5.150 in.
(1.0)(1200)(5.150)
4920
c
1.256 ksi
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500
PROBLEM 4.52
7
8
16 in.
-in. diameter
2 in.
A concrete beam is reinforced by three steel rods placed as shown. The
modulus of elasticity is 3 106 psi for the concrete and 29 106 psi for the
steel. Using an allowable stress of 1350 psi for the concrete and 20 ksi for the
steel, determine the largest allowable positive bending moment in the beam.
8 in.
SOLUTION
As
8x
Locate the neutral axis:
29 10 6
3 106
Es
Ec
n
3
x
2
4
d2
(3)
9.67
4
2
7
8
1.8040 in 2
nAs
17.438 in 2
(17.438)(14 x) 0
4 x 2 17.438 x 244.14 0
Solve for x.
17.438
x
17.4382 (4)(4)(244.14)
(2)(4)
5.6326 in.
14 x 8.3674 in.
I
1 3
8x
3
nMy
I
nAs (14
M
x) 2
1
(8)(5.6326)3
3
(17.438)(8.3674)2
1697.45 in 4
I
ny
n 1.0,
Concrete:
M
Steel:
n
M
Choose the smaller value.
5.6326 in.,
y
(1350)(1697.45)
(1.0)(5.6326)
9.67,
y
(20 103 )(1697.45)
(9.67)(8.3674)
M
1350 psi
406.835 103 lb in.
8.3674 in.,
419.72 lb in.
407 kip in.
407 kip in.
20 103 psi
420 kip in.
M
33.9 kip ft
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501
PROBLEM 4.53
d
The design of a reinforced concrete beam is said to be balanced if the maximum stresses
in the steel and concrete are equal, respectively, to the allowable stresses s and c .
Show that to achieve a balanced design the distance x from the top of the beam to the
neutral axis must be
d
x
1
b
s Ec
c Es
where Ec and Es are the moduli of elasticity of concrete and steel, respectively, and d is
the distance from the top of the beam to the reinforcing steel.
SOLUTION
s
s
c
d
x
x
nM (d x)
Mx
c
I
I
n(d x)
d
n
n
x
x
1
1
n
d
Ec
1
Es
s
1
c
Ec
Es
s
c
s
c
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502
PROBLEM 4.54
d
For the concrete beam shown, the modulus of elasticity is 25 GPa for the concrete and
200 GPa for the steel. Knowing that b = 200 mm and d = 450 mm, and using an
allowable stress of 12.5 MPa for the concrete and 140 MPa for the steel, determine
(a) the required area As of the steel reinforcement if the beam is to be balanced, (b) the
largest allowable bending moment. (See Prob. 4.53 for definition of a balanced beam.)
b
SOLUTION
n
s
s
c
d
x
1
x
0.41667d
bx
Locate neutral axis.
As
(a)
x
2
M
1
c
nAs (d
1 140 106
2.40
8.0 12.5 106
(0.41667)(450) 187.5 mm
x)
(200)(187.5) 2
1674 mm 2
(2)(8.0)(262.5)
M
y 187.5 mm
0.1875 m
(1.3623 10 3 )(12.5 106 )
(1.0)(0.1875)
n 8.0
Steel:
s
bx 2
2n ( d x )
n 1.0
Concrete:
1
n
1 3
1
(200)(187.5)3
bx nAs (d x)2
3
3
9
4
1.3623 10 mm 1.3623 10 3 m 4
nMy
I
M
I
ny
I
(b)
200 109
8.0
25 109
nM (d x)
Mx
c
I
I
n( d x )
d
n
n
x
x
Es
Ec
y
262.5 mm
As
1674 mm 2
(8.0)(1674)(262.5) 2
12.5 106 Pa
90.8 103 N m
0.2625 m
(1.3623 10 3 )(140 106 )
(8.0)(0.2625)
140 106 Pa
90.8 103 N m
Note that both values are the same for balanced design.
M
90.8 kN m
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503
Aluminum
0.5 in.
Brass
0.5 in.
Steel
0.5 in.
Brass
0.5 in.
Aluminum
0.5 in.
1.5 in.
PROBLEM 4.55
Five metal strips, each a 0.5 × 1.5-in. cross section, are bonded together to
form the composite beam shown. The modulus of elasticity is 30 × 106 psi
for the steel, 15 × 106 psi for the brass, and 10 × 106 psi for the aluminum.
Knowing that the beam is bent about a horizontal axis by a couple of
moment 12 kip in., determine (a) the maximum stress in each of the
three metals, (b) the radius of curvature of the composite beam.
SOLUTION
Use aluminum as the reference material.
n
30 106
10 106
Es
Ea
3.0 in steel
15 106
1.5 in brass
10 106
n 1.0 in aluminum
n
Eb
Ea
For the transformed section,
I1
I2
I3
I4
n1
b1h13 n1 A1d12
12
0.7656 in 4
1
(1.5)(0.5)3
12
(0.75)(1.0) 2
n2
1.5
(1.5)(0.5)3 (1.5)(0.75)(0.5)2
b2 h23 n2 A2 d 22
12
12
n3
3.0
(1.5)(0.5)3 0.0469 in 4
b3 h33
12
12
I 2 0.3047 in 4
I 5 I1 0.7656 in 4
5
I
Ii
0.3047 in 4
2.1875 in 4
1
(a)
Aluminum:
nMy
I
(1.0)(12)(1.25)
2.1875
6.86 ksi
Brass:
nMy
I
(1.5)(12)(0.75)
2.1875
6.17 ksi
Steel:
nMy
I
(3.0)(12)(0.25)
2.1875
4.11 ksi
(b)
1
M
Ea I
12 103
(10 106 )(2.1875)
548.57 10
6
in.
1
1823 in. = 151.9 ft
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504
Steel
0.5 in.
Aluminum
0.5 in.
Brass
0.5 in.
Aluminum
0.5 in.
Steel
0.5 in.
1.5 in.
PROBLEM 4.56
Five metal strips, each a 0.5 × 1.5-in. cross section, are bonded together to
form the composite beam shown. The modulus of elasticity is 30 × 106 psi
for the steel, 15 × 106 psi for the brass, and 10 × 106 psi for the aluminum.
Knowing that the beam is bent about a horizontal axis by a couple of
moment 12 kip · in., determine (a) the maximum stress in each of the
three metals, (b) the radius of curvature of the composite beam.
SOLUTION
Use aluminum as the reference material.
n
30 106
10 106
Es
Ea
3.0 in steel
15 106
1.5 in brass
10 106
n 1.0 in aluminum
n
Eb
Ea
For the transformed section,
I1
I2
I3
I4
n1
b1h13 n1 A1d12
12
3.0
(1.5)(0.5)3 (3.0)(0.75)(1.0)2
12
2.2969 in 4
n2
1.0
(1.5)(0.5)3 (1.0)(0.75)(0.5)2
b2 h23 n2 A2 d 22
12
12
n3
1.5
(1.5)(0.5)3 0.0234 in 4
b3 h33
12
12
I 2 0.2031 in 4
I 5 I1 2.2969 in 4
5
I
Ii
0.2031 in 4
5.0234 in 4
1
(a)
Steel:
nMy
I
(3.0)(12)(1.25)
5.0234
Aluminum:
nMy
I
(1.0)(12)(0.75)
1.792 ksi
5.0234
Brass:
nMy
I
(1.5)(12)(0.25)
5.0234
(b)
1
M
Ea I
8.96 ksi
12 103
(10 106 )(5.0234)
0.896 ksi
238.89 10
6
in.
1
4186 in. 349 ft
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505
Brass
PROBLEM 4.57
Aluminum
The composite beam shown is formed by bonding together a brass rod and an
aluminum rod of semicircular cross sections. The modulus of elasticity is
15 106 psi for the brass and 10 106 psi for the aluminum. Knowing that the
composite beam is bent about a horizontal axis by couples of moment
8 kip in., determine the maximum stress (a) in the brass, (b) in the aluminum.
0.8 in.
SOLUTION
r
For each semicircle,
y0
4r
3
I
I base
(4)(0.8)
3
Ay02
0.8 in.
A
r2
2
0.33953 in.
1.00531 in 2 ,
I base
8
r4
0.160850 in 4
(1.00531)(0.33953)2
0.160850
0.044953 in 4
Use aluminum as the reference material.
n 1.0 in aluminum
n
15 106
10 106
Eb
Ea
1.5 in brass
Locate the neutral axis.
A, in2
nA, in2
1.00531
1.50796
0.33953
0.51200
1.00531
1.00531
0.33953
0.34133
y0 , in.
nAy0 , in 3
2.51327
d1
0.06791
I1
n1I
n1 Ad12
I2
n2 I
n2 Ad 22
I
(a)
0.33953
I1
I2
0.27162 in., d 2
(1.5)(0.044957)
0.33953
0.06791
(1.5)(1.00531)(0.27162)
(1.0)(0.044957)
0.17067
2.51327
0.06791 in.
The neutral axis lies 0.06791 in.
above the material interface.
0.17067
2
(1.0)(1.00531)(0.40744)2
0.40744 in.
0.17869 in 4
0.21185 in 4
0.39054 in 4
n
Brass:
1.5,
y
0.8
nMy
I
(b)
Y0
Aluminium:
n
1.0,
y
0.8
nMy
I
0.06791
0.73209 in.
(1.5)(8)(0.73209)
0.39054
0.06791
22.5 ksi
0.86791 in.
(1.0)(8)( 0.86791)
0.39054
17.78 ksi
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506
y
Aluminum
PROBLEM 4.58
3 mm
A steel pipe and an aluminum pipe are securely bonded together to form
the composite beam shown. The modulus of elasticity is 200 GPa for the
steel and 70 GPa for the aluminum. Knowing that the composite beam is
bent by a couple of moment 500 N m, determine the maximum stress
(a) in the aluminum, (b) in the steel.
Steel
6 mm
z
10 mm
38 mm
SOLUTION
Use aluminum as the reference material.
n 1.0 in aluminum
n
Es / Ea
Steel:
Is
ns
Aluminium:
Ia
na
I
Is
200 / 70 2.857 in steel
For the transformed section,
(a)
Aluminum:
4
ro4
ri4
(2.857)
ro4
ri4
(1.0)
4
Ia
c 19 mm
a
na Mc
I
4
4
(164 104 ) 124.62 103 mm 4
(194 164 ) 50.88 103 mm 4
175.50 103 mm 4
175.5 10
9
m4
0.019 m
(1.0)(500)(0.019)
175.5 10 9
54.1 106 Pa
a
(b)
Steel:
c 16 mm
s
ns Mc
I
54.1 MPa
0.016 m
(2.857)(500)(0.016)
175.5 10 9
130.2 106 Pa
s
130.2 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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507
%"
M
PROBLEM 4.59
Et #
100 mm
1
2
Ec
%'
50 mm
Ec
The rectangular beam shown is made of a plastic for which the
value of the modulus of elasticity in tension is one-half of its
value in compression. For a bending moment M 600 N m,
determine the maximum (a) tensile stress, (b) compressive stress.
SOLUTION
n
1
on the tension side of neutral axis
2
n 1 on the compression side
Locate neutral axis.
n1bx
x2
x
h x
(a)
Tensile stress:
x
h x
n2 b(h x)
2
2
1 2 1
bx
b( h x ) 2
2
4
Compressive stress:
0
1
1
(h x) 2 x
(h x)
2
2
1
h 0.41421 h 41.421 mm
2 1
58.579 mm
I1
1
n1 bx3
3
I2
1
n2 b(h x)3
3
I
I1
I2
n
1
,
2
y
nMy
I
(b)
0
n 1,
y
nMy
I
(1)
1
(50)(41.421)3 1.1844 106 mm 4
3
1
2
1
(50)(58.579)3 1.6751 106 mm 4
3
2.8595 106 mm 4
58.579 mm
2.8595 10 6 m 4
0.058579 m
(0.5)(600)( 0.058579)
2.8595 10 6
41.421 mm
6.15 106 Pa
t
6.15 MPa
0.041421 m
(1.0)(600)(0.041421)
2.8595 10 6
8.69 106 Pa
c
8.69 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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508
PROBLEM 4.60*
A rectangular beam is made of material for which the modulus of elasticity is Et in tension and Ec in
compression. Show that the curvature of the beam in pure bending is
1
M
Er I
where
Er
4 Et Ec
Ec ) 2
( Et
SOLUTION
Use Et as the reference modulus.
Then Ec
nEt .
Locate neutral axis.
nbx
nx 2
x
I trans
x
2
b( h x )
(h x) 2
h
nh
h x
n 1
n 3
bx
3
h x
0
2
0
nx (h x)
n 1
1
b ( h x )3
3
M
Et I trans
Er I
Et I trans
Er
Et I trans
I
where I
12
bh3
Et
1
n 1
1
3
bh3
n
n 1
2
bh3
1 3
bh
12
n
3
3
n
n 1
M
Er I
4 Et Ec /Et
Ec /Et
3
1
n
1n 1
bh3
3 n 13
1 n n3/ 2
bh3
3 n 13
1
h
3
n 1
2
bh3
4 Et Ec
2
Ec
Et
2
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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509
PROBLEM 4.61
8 mm
r
M
80 mm
Knowing that M 250 N m, determine the maximum stress in the beam
shown when the radius r of the fillets is (a) 4 mm, (b) 8 mm.
40 mm
SOLUTION
c
1 3
bh
12
20 mm
D
d
80 mm
40 mm
2.00
r
d
4 mm
40 mm
0.10
I
(a)
max
(b)
r
d
K
Mc
I
8 mm
40 mm
max
K
1
(8)(40)3
12
0.020 m
42.667 103 mm 4
0.20
Mc
I
K
From Fig. 4.27,
(1.87)(250)(0.020)
42.667 10 9
K
176.0 106 Pa
9
m4
1.87
219 106 Pa
From Fig. 4.27,
(1.50)(250)(0.020)
42.667 10 9
42.667 10
max
219 MPa
1.50
max
176.0 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
510
PROBLEM 4.62
8 mm
r
M
80 mm
Knowing that the allowable stress for the beam shown is 90 MPa, determine
the allowable bending moment M when the radius r of the fillets is (a) 8 mm,
(b) 12 mm.
40 mm
SOLUTION
c
1 3
bh
12
20 mm
D
d
80 mm
40 mm
2.00
r
d
8 mm
40 mm
0.2
I
(a)
max
(b)
r
d
K
12 mm
40 mm
1
(8)(40)3
12
0.020 m
Mc
I
0.3
42.667 103 mm 4
K
From Fig. 4.27,
M
max I
Kc
9
m4
1.50
(90 106 )(42.667 10 9 )
(1.50)(0.020)
K
From Fig. 4.27,
M
42.667 10
M
128.0 N m
M
142.0 N m
1.35
(90 106 )(42.667 10 9 )
(1.35)(0.020)
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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511
3
4
r
PROBLEM 4.63
in.
Semicircular grooves of radius r must be milled as shown in the sides of
a steel member. Using an allowable stress of 8 ksi, determine the largest
bending moment that can be applied to the member when (a) r = 83 in.,
(b) r = 34 in.
M
4.5 in.
SOLUTION
(a)
d
D
2r
D
d
4.5
3.75
4.5
1.20
From Fig. 4.28,
I
1 3
bh
12
Mc
K
I
(2)
3
8
r
d
3.75 in.
0.375
3.75
K
0.10
2.07
1 3
(3.75)3
12 4
I
M
Kc
3.296 in 4
1
2
c
(8)(3.296)
(2.07)(1.875)
1.875 in.
6.79 kip in.
6.79 kip in.
(b)
d
D
2r
4.5
From Fig. 4.28,
(2)
3
4
3.0
D
d
4.5
3.0
K
1.61
I
c
1
d
2
1.5 in.
r
d
1.5
1 3
bh
12
M
0.75
3.0
1 3
(3.0)3
12 4
I
Kc
0.25
1.6875 in 4
(8)(1.6875)
(1.61)(1.5)
5.59 kip in.
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on a website, in whole or part.
512
3
4
r
PROBLEM 4.64
in.
Semicircular grooves of radius r must be milled as shown in the sides
of a steel member. Knowing that M = 4 kip · in., determine the
maximum stress in the member when the radius r of the semicircular
grooves is (a) r = 83 in., (b) r = 34 in.
M
4.5 in.
SOLUTION
(a)
From Fig. 4.28,
d
D
D
d
4.5
3.75
K
2.07
2r
4.5
r
d
1.20
1 3
bh
12
Mc
K
I
I
3
8
(2)
3.75 in.
0.375
3.75
0.10
1 3
(3.75)3 3.296 in 4
12 4
(2.07)(4)(1.875)
4.71 ksi
3.296
c
1
2
1.875 in.
4.71 ksi
(b)
d
D
2r
4.5
(2)
3
4
3.00 in.
D
d
4.5
3.00
1.50
r
d
0.75
3.0
c
1
d
2
1.5 in.
0.25
From Fig. 4.28,
K
1.61
I
1 3
bh
12
1 3
(3.00)3
12 4
K
Mc
I
1.6875 in 4
(1.61)(4)(1.5)
1.6875
5.72 ksi
5.72 ksi
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513
M
PROBLEM 4.65
M
100 mm
100 mm
150 mm
150 mm
18 mm
(a)
18 mm
A couple of moment M = 2 kN · m is to be
applied to the end of a steel bar. Determine the
maximum stress in the bar (a) if the bar is
designed with grooves having semicircular
portions of radius r = 10 mm, as shown in Fig. a,
(b) if the bar is redesigned by removing the
material to the left and right of the dashed lines
as shown in Fig. b.
(b)
SOLUTION
For both configurations,
D
150 mm
d
100 mm
r
10 mm
D
d
r
d
150
100
10
100
Fig. 4.28 gives
Ka
2.21.
For configuration (b),
Fig. 4.27 gives K b
1.50
0.10
For configuration (a),
I
c
(a)
KMc
I
1.79.
1 3
1
(18)(100)3 1.5 106 mm 4
bh
12
12
1
d 50 mm 0.05 m
2
(2.21)(2 103 )(0.05)
1.5 10 6
147.0 106 Pa
1.5 10
6
m4
147.0 MPa
147.0 MPa
(b)
KMc
I
(1.79)(2 103 )(0.05)
1.5 10 6
119.0 106 Pa
119.0 MPa
119.0 MPa
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514
M
100 mm
150 mm
PROBLEM 4.66
M
100 mm
150 mm
18 mm
(a)
18 mm
The allowable stress used in the design of a
steel bar is 80 MPa. Determine the largest
couple M that can be applied to the bar (a) if
the bar is designed with grooves having
semicircular portions of radius r 15 mm, as
shown in Fig. a, (b) if the bar is redesigned by
removing the material to the left and right of
the dashed lines as shown in Fig. b.
(b)
SOLUTION
For both configurations,
D
150 mm
r
15 mm
D
d
r
d
150
100
15
100
d
100 mm
1.50
0.15
For configuration (a), Fig. 4.28 gives K a
1.92.
For configuration (b), Fig. 4.27 gives Kb
1.57.
I
c
KMc
I
(a)
M
I
Kc
1 3
1
(18)(100)3 1.5 106 mm 4
bh
12
12
1
d 50 mm 0.050 m
2
(80 106 )(1.5 10 6 )
(1.92) (0.05)
1.5 10
6
m4
1.250 103 N m
1.250 kN m
(a)
M
I
Kc
(80 106 )(1.5 10 6 )
(1.57)(0.050)
1.530 103 N m
1.530 kN m
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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515
M
PROBLEM 4.67
The prismatic bar shown is made of a steel that is assumed to be
elastoplastic with Y 300 MPa and is subjected to a couple M
parallel to the x axis. Determine the moment M of the couple for which
(a) yield first occurs, (b) the elastic core of the bar is 4 mm thick.
x
z
12 mm
8 mm
SOLUTION
(a)
I
1 3
bh
12
1
(12)(8)3
12
c
yY
1
(4)
2
2 mm
512 10 12 m 4
1
h 4 mm
0.004 m
2
(300 106 )(512 10
c
0.004
38.4 N m
YI
MY
(b)
512 mm 4
yY
c
2
4
M
3
1 yY
MY 1
2
3 c
12
)
MY
38.4 N m
M
52.8 N m
0.5
2
3
1
(38.4) 1
(0.5) 2
2
3
52.8 N m
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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516
M
PROBLEM 4.68
Solve Prob. 4.67, assuming that the couple M is parallel to the z axis.
PROBLEM 4.67 The prismatic bar shown is made of a steel that is
assumed to be elastoplastic with Y 300 MPa and is subjected to a
couple M parallel to the x axis. Determine the moment M of the couple
for which (a) yield first occurs, (b) the elastic core of the bar is 4 mm
thick.
x
z
12 mm
8 mm
SOLUTION
(a)
I
1 3
bh
12
1
(8)(12)3
12
c
yY
1
(4)
2
2 mm
1.152 10 9 m 4
1
h 6 mm 0.006 m
2
(300 106 )(1.152 10 9 )
c
0.006
57.6 N m
YI
MY
(b)
1.152 103 mm 4
yY
c
2
6
M
3
1 yY
MY 1
2
3 c
MY
57.6 N m
M
83.2 N m
1
3
3
1 1
(57.6) 1
2
3 3
2
2
83.2 N m
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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517
PROBLEM 4.69
A solid square rod of side 0.6 in. is made of a steel that is assumed to be elastoplastic with E = 29 × 106 psi
and Y 48 ksi. Knowing that a couple M is applied and maintained about an axis parallel to a side of the
cross section, determine the moment M of the couple for which the radius of curvature is 6 ft.
SOLUTION
I
MY
1
(0.6)(0.6)3
12
I
Y
c
c
yY
c
M
Y
M
3
in 4
4.16667 10
1.65517 10
4.16667 10
3
MY 1
2
1 YY
3 c
2
3
1
h
2
c
(10.8 10 3 )(48 103 )
0.3
0.3
72
max
10.8 10
0.3 in.
1728 lb in.
Y
Y
E
48 103
29 106
6 ft
72 in.
1.65517 10
3
3
3
0.39724
3
(1728) 1
2
1
(0.39724)2
3
M
2460 lb in.
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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518
PROBLEM 4.70
For the solid square rod of Prob. 4.69, determine the moment M for which the radius of curvature is 3 ft.
PROBLEM 4.69. A solid square rod of side 0.6 in. is made of a steel that is assumed to be elastoplastic with
E = 29 × 106 psi and Y 48 ksi. Knowing that a couple M is applied and maintained about an axis parallel
to a side of the cross section, determine the moment M of the couple for which the radius of curvature is 6 ft.
SOLUTION
I
MY
1
(0.6)(0.6)3
12
I
Y
c
yY
c
M
3
in 4
Y
max
8.3333 10
1.65517 10 3
8.3333 10 3
3
MY 1
2
1 YY
3 c
2
3
1
h
2
c
(10.8 10 3 )(48 103 )
0.3
0.3
36
c
max
10.8 10
0.3 in.
1728 lb in.
Y
48 103
29 106
Y
E
3 ft
36 in.
1.65517 10
3
0.19862
3
(1728) 1
2
1
(0.19862)2
3
M
2560 lb in.
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
519
y
PROBLEM 4.71
18 mm
The prismatic rod shown is made of a steel that is assumed to be elastoplastic with
E 200 GPa and Y 280 MPa . Knowing that couples M and M of moment
525 N · m are applied and maintained about axes parallel to the y axis, determine
(a) the thickness of the elastic core, (b) the radius of curvature of the bar.
M
24 mm
M!
x
SOLUTION
I
c
MY
M
1 3
1
bh
(24)(18)3 11.664 103 mm 4
12
12
1
h 9 mm 0.009 m
2
YI
c
(280 106 )(11.664 10 9 )
0.009
3
MY 1
2
yY
c
3
1 yY 2
3 c2
(2)(525)
362.88
or
0.32632
yY
c
yY
11.664 10
3
m4
362.88 N m
3
2
M
MY
0.32632c
2.9368 mm
(a)
(b)
tcore
Y
yY
Y
E
EyY
Y
(200 109 )(2.9368 10 3 )
280 106
2 yY
5.87 mm
2.09 m
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520
y
PROBLEM 4.72
18 mm
Solve Prob. 4.71, assuming that the couples M and M are applied and
maintained about axes parallel to the x axis.
M
24 mm
PROBLEM 4.71 The prismatic rod shown is made of a steel that is assumed to
be elastoplastic with E 200 GPa and Y 280 MPa . Knowing that couples
M and M of moment 525 N · m are applied and maintained about axes parallel
to the y axis, determine (a) the thickness of the elastic core, (b) the radius of
curvature of the bar.
M!
x
SOLUTION
I
c
MY
M
yY
c
1 3
1
bh
(18)(24)3 20.736 103 mm 4
12
12
1
h 12 mm 0.012 m
2
YI
c
(280 106 )(20.736 10 9 )
0.012
3
MY 1
2
3
1 yY 2
3 c2
(2)(525)
483.84
or
yY
c
0.91097
yY
20.736 10
9
m4
483.84 N m
3
2
M
MY
0.91097
c
10.932 mm
(a)
(b)
tcore
Y
yY
Y
E
EyY
Y
(200 109 )(10.932 10 3 )
280 106
2 yY
21.9 mm
7.81 m
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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521
y
z
PROBLEM 4.73
C
90 mm
A beam of the cross section shown is made of a steel that is assumed to be
elastoplastic with E 200 GPa and Y 240 MPa. For bending about the z axis,
determine the bending moment at which (a) yield first occurs, (b) the plastic
zones at the top and bottom of the bar are 30 mm thick.
60 mm
SOLUTION
(a)
I
c
MY
1 3 1
(60)(90)3 3.645 106 mm 4 3.645 10 6 m 4
bh
12
12
1
h 45 mm 0.045 m
2
(240 106 )(3.645 10 6 )
YI
19.44 10 N m
0.045
c
MY
R1
Y
19.44 kN m
(240 106 )(0.060)(0.030)
A1
432 103 N
y1 15 mm 15 mm
R2
1
2
Y
A2
0.030 m
1
(240 106 )(0.060)(0.015)
2
108 103 N
y2
(b)
M
2( R1 y1
R2 y2 )
2
(15 mm) 10 mm
3
0.010 m
2[(432 103 )(0.030) (108 103 )(0.010)]
28.08 103 N m
M
28.1 kN m
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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522
PROBLEM 4.74
y
A beam of the cross section shown is made of a steel that is assumed to be
elastoplastic with E 200 GPa and Y 240 MPa. For bending about the
z axis, determine the bending moment at which (a) yield first occurs,
(b) the plastic zones at the top and bottom of the bar are 30 mm thick.
30 mm
30 mm
C
z
30 mm
15 mm
30 mm
15 mm
SOLUTION
(a)
1 3 1
bh
(60)(90)3 3.645 106 mm 4
12
12
1 3 1
I cutout
bh
(30)(30)3 67.5 103 mm 4
12
12
I 3.645 106 67.5 103 3.5775 106 mm 4
I rect
3.5775 10
c
MY
45 mm
y1
15 mm
M
0.045 m
(240 106 )(3.5775 10 6 )
0.045
c
3
19.08 10 N m
Y A1
y2
mm 4
YI
R1
R2
(b)
1
h
2
6
(240 106 )(0.060)(0.030)
15 mm
30 mm
19.08 kN m
M
27.0 kN m
432 103 N
0.030 m
1
1
(240 106 )(0.030)(0.015)
Y A2
2
2
2
(15 mm) 10 mm 0.010 m
3
2( R1 y1
MY
54 103 N
R2 y2 )
2[(432 103 )(0.030)
(54 103 )(0.010)]
27.00 103 N m
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523
PROBLEM 4.75
y
3 in.
3 in.
C
z
A beam of the cross section shown is made of a steel that is assumed to
be elastoplastic with E 29 106 psi and Y 42 ksi. For bending
about the z axis, determine the bending moment at which (a) yield first
occurs, (b) the plastic zones at the top and bottom of the bar are 3 in.
thick.
3 in.
1.5 in.
1.5 in.
3 in.
SOLUTION
(a)
I1
I2
I3
1
1
b1h13 A1d12
(3)(3)3 (3)(3)(3) 2
12
12
1
1
b2 h23
(6)(3)3 13.5 in 4
12
12
I1 87.75 in 4
I 3 188.5 in 4
I
I1
c
4.5 in.
(42)(188.5)
YI
c
4.5
MY
I2
87.75 in 4
MY
R1
Y
A1
1759 kip in.
(42)(3)(3) 378 kip
y1 1.5 1.5 3.0 in.
R2
y2
(b)
M
2( R1 y1
R2 y2 )
1
1
(42)(6)(1.5)
Y A2
2
2
189 kip
2
(1.5) 1.0 in.
3
2[(378)(3.0) (189)(1.0)] 2646 kip in.
M
2650 kip in.
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
524
PROBLEM 4.76
y
3 in.
3 in.
C
z
A beam of the cross section shown is made of a steel that is assumed to
be elastoplastic with E 29 106 psi and Y 42 ksi. For bending
about the z axis, determine the bending moment at which (a) yield first
occurs, (b) the plastic zones at the top and bottom of the bar are 3 in.
thick.
3 in.
1.5 in.
3 in.
1.5 in.
SOLUTION
(a)
I1
I2
I3
1
1
(6)(3)3 (6)(3)(3)2
b1h13 A1d12
12
12
1
1
3
(3)(3)3 6.75 in 4
b2 h2
12
12
I1 175.5 in 4
I
I1
c
4.5 in.
MY
I2
YI
c
I3
175.5 in 4
357.75 in 4
(42)(357.75)
4.5
3339 kip in.
MY
R1
Y
(42)(6)(3)
A1
3340 kip in.
756 kip
y1 1.5 1.5 3 in.
(b)
M
2( R1 y1
R2 y2 )
R2
1
2
y2
2
(1.5) 1.0 in.
3
2[(756)(3) (94.5)(1.0)] 4725 kip in.
Y
A2
1
(42)(3)(1.5)
2
94.5 kip
M 4730 kip in.
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
525
PROBLEM 4.77
y
For the beam indicated (of Prob. 4.73), determine (a) the fully plastic moment
M p , (b) the shape factor of the cross section.
z
90 mm
C
60 mm
SOLUTION
From Problem 4.73,
E
200 GPa and
A1
(60)(45)
R
2700 10 6 m 2
Y A1
Y
240 MPa
2700 mm 2
(240 106 )(2700 10 6 )
d
(a)
Mp
(b)
I
c
MY
Rd
(648 103 )(0.045)
1 3 1
bh
(60)(90)3
12
12
45 mm 0.045 m
YI
c
648 103 N
45 mm 0.045 m
29.16 103 N m
3.645 106 mm 4
(240 106 )(3.645 10 6 )
0.045
k
Mp
29.2 kN m
3.645 10 6 m 4
19.44 103 N m
Mp
MY
29.16
19.44
k 1.500
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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526
PROBLEM 4.78
y
For the beam indicated (of Prob. 4.74), determine (a) the fully plastic moment
M p , (b) the shape factor of the cross section.
30 mm
30 mm
C
z
30 mm
15 mm
15 mm
30 mm
SOLUTION
From Problem 4.74,
(a)
R1
y1
Y
200 GPa and
Y
240 MPa
A1
(240 106 )(0.060)(0.030)
432 103 N
15 mm 15 mm 30 mm
0.030 m
R2
y2
E
Y
A2
(240 106 )(0.030)(0.015)
108 103 N
1
(15) 7.5 mm 0.0075 m
2
2( R1 y1
Mp
2[( 432 103 )(0.030) (108 103 )(0.0075)]
R2 y2 )
27.54 103 N m
(b)
I rect
I cutout
I
Mp
1 3 1
(60)(90)3 3.645 106 mm 4
bh
12
12
1 3 1
bh
(30)(30)3 67.5 103 mm 4
12
12
I rect I cutout 3.645 106 67.5 103 3.5775 103 mm 4
3.5775 10
c
MY
k
27.5 kN m
1
h
2
YI
c
Mp
MY
45 mm
9
m4
0.045 m
(240 106 )(3.5775 10 9 )
0.045
27.54
19.08
19.08 103 N m
k 1.443
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
527
PROBLEM 4.79
y
3 in.
3 in.
C
z
For the beam indicated (of Prob. 4.75), determine (a) the fully plastic
moment M p , (b) the shape factor of the cross section.
3 in.
1.5 in.
1.5 in.
3 in.
SOLUTION
From Problem 4.75,
(a)
R1
Y
A1
E
29 106 psi and
Y
42 ksi.
(42)(3)(3) 378 kip
y1 1.5 1.5 3.0 in.
R2
(b)
Y
A2
(42)(6)(1.5) 378 kip
y2
1
(1.5)
2
0.75 in.
Mp
2( R1 y1
R2 y2 )
I1
I2
I3
1
1
b1h13 A1d12
(3)(3)3 (3)(3)(3) 2
12
12
1
1
b2 h23
(6)(3)3 13.5 in 4
12
12
I1 87.75 in 4
I
I1
c
4.5 in.
MY
k
2[( 378)( 3.0) ( 378)(0. 75)]
YI
c
Mp
MY
I2
Mp
2840 kip in.
87.75 in 4
I 3 188.5 in 4
(42)(188.5)
4.5
2835
1759.3
1759.3 kip in
k 1.611
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
528
PROBLEM 4.80
y
For the beam indicated (of Prob. 4.76), determine (a) the fully plastic
moment M p , (b) the shape factor of the cross section.
3 in.
3 in.
C
z
3 in.
1.5 in.
3 in.
1.5 in.
SOLUTION
From Problem 4.76,
(a)
R1
Y
A1
E
(42)(6)(3)
29 106 and
Y
42 ksi
756 kip
y1 1.5 1.5 3.0 in.
R2
(b)
Y
A2
(42)(3)(1.5) 189 kip
y2
1
(1.5)
2
0.75 in.
Mp
2( R1 y1
R2 y2 )
I1
I2
I3
1
1
b1h13 A1d12
(6)(3)3 (6)(3)(3)2
12
12
1
1
3
b2 h2
(3)(3)3 6.75 in 4
12
12
I1 175.5 in 4
I
I1
c
4.5 in.
MY
k
2[( 756)( 3.0) (189)(0. 75)]
I2
YI
c
Mp
MY
I3
Mp
4820 kip in.
175.5 in 4
357.75 in 4
(42)(357.75)
4.5
4819.5
3339
3339 kip in.
k 1.443
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
529
PROBLEM 4.81
r ! 18 mm
Determine the plastic moment M p of a steel beam of the cross section shown,
assuming the steel to be elastoplastic with a yield strength of 240 MPa.
SOLUTION
For a semicircle,
A
2
Resultant force on semicircular section:
r 2;
R
r
4r
3
YA
Resultant moment on entire cross section:
Mp
Data:
Y
Mp
2 Rr
4
3
240 MPa
Yr
3
240 106 Pa,
4
(240 106 )(0.018)3
3
r
18 mm
0.018 m
1866 N m
Mp
1.866 kN m
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
530
PROBLEM 4.82
50 mm
30 mm
Determine the plastic moment M p of a steel beam of the cross section shown,
assuming the steel to be elastoplastic with a yield strength of 240 MPa.
10 mm
10 mm
30 mm
10 mm
SOLUTION
A (50)(90) (30)(30) 3600 mm 2
Total area:
1
A 1800 mm 2
2
1
A 1800
x 2
50
b
A1
(50)(36) 1800 mm 2 ,
A2
(50)(14)
A3
(20)(30)
A4
36 mm
y1 18 mm,
A1 y1
32.4 103 mm3
700 mm 2 ,
y2
7 mm,
A2 y2
4.9 103 mm3
600 mm 2 ,
y3
29 mm,
A3 y3
17.4 103 mm3
(50)(10) 500 mm 2 ,
y4
49 mm,
A4 y4
24.5 103 mm3
A1 y1
Mp
A2 y2
Y
Ai yi
A3 y3
A4 y4
79.2 103 mm3
79.2 10 6 m3
(240 106 )(79.2 10 6 ) 19.008 103 N m
Mp
19.01 kN m
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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531
PROBLEM 4.83
36 mm
Determine the plastic moment M p of a steel beam of the cross section shown,
assuming the steel to be elastoplastic with a yield strength of 240 MPa.
30 mm
SOLUTION
Total area:
A
1
(30)(36) 540 mm 2
2
1
A 270 mm 2
2
Half area:
By similar triangles,
b
y
30
36
b
Since
A1
1
by
2
5 2
y ,
12
A1
5
y
6
y2
12
A1
5
12
(270) 25.4558 mm
5
b 21.2132 mm
1
(21.2132)(25.4558) 270 mm 2 270 10 6 m 2
A1
2
A2 (21.2132)(36 25.4558) 223.676 mm 2 223.676 10 6 m 2
y
A3
Ri
R1
y1
y2
y3
Mp
A
Y
A1
Ai
A2
46.324 mm 2
46.324 10 6 m 2
240 106 Ai
64.8 103 N, R2 53.6822 103 N, R3 11.1178 103 N
1
y 8.4853 mm 8.4853 10 3 m
3
1
(36 25.4558) 5.2721 mm 5.2721 10 3 m
2
2
(36 25.4558) 7.0295 mm 7.0295 10 3 m
3
R1 y1 R2 y2 R3 y3 911 N m
Mp
911 N m
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532
0.4 in.
PROBLEM 4.84
1.0 in.
Determine the plastic moment Mp of a steel beam of the cross section shown,
assuming the steel to be elastoplastic with a yield strength of 42 ksi.
1.0 in.
0.4 in.
0.4 in.
SOLUTION
A (1.0 in.)(0.4 in.) 2(1.4 in.)(0.4 in.) 1.52 in 2
x
R1
1
2
(1.52)
0.95 in.
2(0.4)
0.4 in 2
(1.0)(0.4)
y1 1.2 0.95 0.25 in.
R2
y2
R3
2(0.4)(1.4 0.95)
0.36 in 2
1
(1.4 0.95) 0.225 in.
2
2(0.4)(0.95) 0.760 in 2
y3
1
(0.95)
2
Mp
( R1 y1
0.475 in.
R2 y 2
R3 y 3 )(
y)
[(0.4)(0.25) (0.36)(0.225) (0.760)(0.475)](42)
Mp
22.8 kip in .
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533
5 mm
PROBLEM 4.85
5 mm
80 mm
t = 5 mm
120 mm
Determine the plastic moment Mp of the cross section shown when the beam is
bent about a horizontal axis. Assume the material to be elastoplastic with a
yield strength of 175 MPa.
SOLUTION
For M p , the neutral axis divides the area into two equal parts.
Total area
(100 100 120)t
320t
1
(320)t
2
a 80 mm
80
b
(80 mm) 64 mm
100
Shaded area
6
2at
m2
A1
2at
A2
2(100 a )t
2(0.02 m)(0.005 m)
200 10
A3
(120 mm)t
(0.120 m)(0.005 m)
600 10
2(0.08 m)(0.005 m) 800 10
R1
Y
A1
(175 MPa)(800 10
6
m 2 ) 140 kN
R2
Y
A2
(175 MPa)(200 10
6
m 2 ) 35 kN
R3
Y
A3
(175 MPa)(600 10
6
m 2 ) 105 kN
Mz : M p
6
6
m2
m2
R1 (32 mm) R2 (8 mm) R3 (16 mm)
(140 kN)(0.032 m) (35 kN)(0.008 m) (105 kN)(0.016 m)
M p = 6.44 k N m
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534
PROBLEM 4.86
4 in.
1
2
in.
1
2
in.
1
2
in.
3 in.
Determine the plastic moment M p of a steel beam of the cross section shown, assuming
the steel to be elastoplastic with a yield strength of 36 ksi.
2 in.
SOLUTION
Total area:
A (4)
1
2
1
1
(3) (2)
2
2
4.5 in 2
1
A 2.25 in 2
2
A1
2.00 in 2 ,
y1
0.75,
A1 y1 1.50 in 3
A2
0.25 in 2 ,
y2
0.25,
A2 y2
0.0625 in 3
A3 1.25 in 2 ,
y3
1.25,
A3 y3
1.5625 in 3
1.00 in 2 ,
y4
2.75,
A4 y4
2.75 in 3
A4
Mp
Y
( A1 y1
A2 y2
A3 y3
A4 y4 )
(36)(1.50 0.0625 1.5625 2.75)
Mp
212 kip in .
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535
PROBLEM 4.87
y
z
C
90 mm
For the beam indicated (of Prob. 4.73), a couple of moment equal to the full
plastic moment M p is applied and then removed. Using a yield strength of
240 MPa, determine the residual stress at y 45 mm .
60 mm
SOLUTION
29.16 103 N m
Mp
See solutions to Problems 4.73 and 4.77.
I
3.645 10 6 m 4
c
0.045 m
M max y
I
LOADING
Mp c
UNLOADING
(29.16 103 )(0.045)
3.645 10 6
res
Y
360 106
120 106 Pa
I
at y
c
45 mm
RESIDUAL STRESSES
360 106 Pa
240 106
res
120.0 MPa
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536
PROBLEM 4.88
y
30 mm
z
C
30 mm
For the beam indicated (of Prob. 4.74), a couple of moment equal to the
full plastic moment M p is applied and then removed. Using a yield
strength of 240 MPa, determine the residual stress at y 45mm .
30 mm
15 mm
30 mm
15 mm
SOLUTION
Mp
I
27.54 103 N m (See solutions to Problems 4.74 and 4.78.)
3.5775 10 6 m 4 ,
M max y
I
Mp c
I
at
c
0.045 m
y
c
(27.54 103 )(0.045)
3.5775 10 6
LOADING
UNLOADING
res
Y
346.4 106
240 106
346.4 106 Pa
RESIDUAL STRESSES
106.4 106 Pa
res
106.4 MPa
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537
PROBLEM 4.89
y
A bending couple is applied to the bar indicated, causing plastic zones
3 in. thick to develop at the top and bottom of the bar. After the couple
has been removed, determine (a) the residual stress at y 4.5 in.,
(b) the points where the residual stress is zero, (c) the radius of
curvature corresponding to the permanent deformation of the bar.
3 in.
3 in.
C
z
Beam of Prob. 4.75.
3 in.
1.5 in.
1.5 in.
3 in.
SOLUTION
See solution to Problem 4.75 for bending couple and stress distribution during loading.
M
2646 kip in.
42 ksi
Y
Mc
I
MyY
I
(a)
At
y
c,
At
y
yY ,
res
(c)
y0
I Y
M
At
y
188.5 in 4
I
63.167
Y
res
Y
c
E
29 106 psi
29 103 ksi
4.5 in.
42
21.056
21.167 ksi
42
20.944 ksi
UNLOADING
My0
Y
I
(188.5)(42)
2646
0
res
1.5 in.
(2646)(4.5)
63.167 ksi
188.5
(2646)(1.5)
21.056 ksi
188.5
LOADING
(b)
yY
yY ,
Ey
2.99 in.
20.9 ksi
res
RESIDUAL STRESSES
Answer:
y0
2.99 in.,
0,
2.99 in.
20.944 ksi
res
Ey
(29 103 )(1.5)
20.944
2077 in.
173.1 ft
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538
PROBLEM 4.90
y
A bending couple is applied to the bar indicated, causing plastic zones
3 in. thick to develop at the top and bottom of the bar. After the couple
has been removed, determine (a) the residual stress at y 4.5 in.,
(b) the points where the residual stress is zero, (c) the radius of
curvature corresponding to the permanent deformation of the bar.
3 in.
3 in.
C
z
Beam of Prob. 4.76.
3 in.
1.5 in.
3 in.
1.5 in.
SOLUTION
See solution to Problem 4.76 for bending couple and stress distribution.
M
Y
4725 kip in.
42 ksi
I
yY
1.5 in.
357.75 in 4
At
At
y
yY ,
res
Y
I Y
M
y0
(c)
At
My0
I
0
y
19.8113
59.4 ksi
17.4340 ksi
42
22.189 ksi
UNLOADING
LOADING
res
29 103 ksi
4.5 in.
Mc (4725)(4.5)
59.434 ksi
I
357.75
MyY
(4725)(1.5)
19.8113 ksi
I
357.75
y c,
59.434 42
res
Y
(a)
(b)
c
29 106 psi
E
0
Y
(357.75)(42)
4725
yY ,
Ey
res
RESIDUAL STRESSES
3.18 in.
Answer:
y0
3.18 in., 0, 3.18 in.
22.189 ksi
Ey
(29 103 )(1.5)
22.189
1960.43 in.
163.4 ft.
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on a website, in whole or part.
539
PROBLEM 4.91
y
z
A bending couple is applied to the beam of Prob. 4.73, causing plastic zones
30 mm thick to develop at the top and bottom of the beam. After the couple has
been removed, determine (a) the residual stress at y 45 mm, (b) the points
where the residual stress is zero, (c) the radius of curvature corresponding to
the permanent deformation of the beam.
90 mm
C
60 mm
SOLUTION
See solution to Problem 4.73 for bending couple and stress distribution during loading.
28.08 103 N m
M
Mc
I
c,
At y
yY ,
res
y0
E
c
(28.08 103 )(0.045)
3.645 10 6
346.67 106 Pa
346.67 MPa
(28.08 103 )(0.015)
3.645 10 6
res
346.67
Y
res
115.556 106 Pa
Y
240
115.556
LOADING
(b)
0.015 m
3.645 10 6 m 4
M yY
I
At y
15 mm
I
240 MPa
Y
(a)
yY
0.045 m
115.556 MPa
106.670 MPa
240
My0
I
I Y
M
(3.645 10 6 )(240 106 )
28.08 103
Y
res
124.444 MPa
UNLOADING
0
200 GPa
res
At
y
yY ,
Ey
res
124.4 MPa
RESIDUAL STRESSES
0
31.15 10 3 m
31.15 mm
Answer: y0
(c)
106.7 MPa
31.2 mm, 0, 31.2 mm
124.444 106 Pa
Ey
(200 109 )(0.015)
124.444 106
24.1 m
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540
PROBLEM 4.92
y
A beam of the cross section shown is made of a steel that is assumed to be
elastoplastic with E 29 × 106 psi and Y = 42 ksi. A bending couple is
applied to the beam about the z axis, causing plastic zones 2 in. thick to
develop at the top and bottom of the beam. After the couple has been removed,
determine (a) the residual stress at y 2 in., (b) the points where the residual
stress is zero, (c) the radius of curvature corresponding to the permanent
deformation of the beam.
1 in.
z
2 in.
C
1 in.
1 in.
1 in.
1 in.
SOLUTION
Flange:
I1
Web:
I2
1
b1h13
12
1
(3)(1)3
12
A1d12
1
1
b2 h23
(1)(2)3
12
12
I1 7.0 in 4
I3
I
I1
I2
c
2 in.
YI
MY
Y
7.0 in 4
0.6667 in 4
14.6667 in 4
(42)(14.6667)
2
c
R1
I3
(3)(1)(1.5)2
A1
MY
308 kip in .
M
406 kip in .
(42)(3)(1) 126 kips
y1 1.0 0.5 1.5 in.
1
Y A2
2
21 kips
R2
M
Y
(a)
y2
2
(1.0) 0.6667 in.
3
M
2( R1 y1
406 kip in. yY
42 ksi
1
(42)(1)(1)
2
I
R2 y 2 )
1.0 in. E
14.6667 in 4 c
Mc (406)(2)
I 14.6667
2[(126)(1.5) (21)(0.6667)]
29 106
29 103 ksi
2 in.
55.364
MyY (406)(1.0)
I
14.662
27.682
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541
PROBLEM 4.92 (Continued)
At y
c,
res
55.364 42 13.3640 ksi
Y
res
At y
yY ,
res
Y
27.682 42
14.3180 ksi
res
(b)
res
y0
0
I Y
M
At y
yY ,
14.32 k si
My0
0
Y
I
(14.667)(42)
1.517 in
406
Answer : y0
(c)
13.36 k si
res
Ey
1.517 in., 0, 1.517 in.
14.3180 ksi
Ey
(29 103 )(1.0)
14.3180
2025.42 in.
168.8 ft
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542
PROBLEM 4.93*
A rectangular bar that is straight and unstressed is bent into an arc of circle of radius by two couples of
moment M. After the couples are removed, it is observed that the radius of curvature of the bar is R .
Denoting by Y the radius of curvature of the bar at the onset of yield, show that the radii of curvature satisfy
the following relation:
1
1
3
2
1
R
2
1
3
1
Y
Y
SOLUTION
1
Y
Let m denote
MY
,
EI
M
3
MY 1
2
1
3
2
2
2
Y
2
Y
2
,
2
Y
M
.
MY
m
M
MY
1
1
M
EI
R
1
3
1
2
1
1
mM Y
EI
1
m
Y
1
3
1
2m
m
Y
3
2
1
Y
1
3
2
2
Y
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543
PROBLEM 4.94
A solid bar of rectangular cross section is made of a material that is assumed to be elastoplastic. Denoting by
M Y and Y , respectively, the bending moment and radius of curvature at the onset of yield, determine (a) the
radius of curvature when a couple of moment M 1.25 M Y is applied to the bar, (b) the radius of curvature
after the couple is removed. Check the results obtained by using the relation derived in Prob. 4.93.
SOLUTION
(a)
1
Y
m
MY
,
EI
M
MY
3
MY 1
2
M
3
1
2
1
3
1
3
2
Let m
2
Y
M
MY
1.25
2
2
Y
3
2m
0.70711
Y
0.707
(b)
1
1
M
EI
R
1
mM Y
EI
1
m
Y
1
0.70711
Y
1.25
Y
Y
0.16421
R
6.09
Y
Y
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544
PROBLEM 4.95
M
The prismatic bar AB is made of a steel that is assumed to be elastoplastic and
for which E 200 GPa . Knowing that the radius of curvature of the bar is
2.4 m when a couple of moment M 350 N m is applied as shown,
determine (a) the yield strength of the steel, (b) the thickness of the elastic
core of the bar.
B
A
20 mm
16 mm
SOLUTION
M
1
3
3 YI
1
2 c
1 2 Y2
3 E 2c 2
3
2
Y b(2c)
bc 2
2
Y
1
2
Y
2 2
3E c
M
Data:
2
Y
3
1 2 Y2
3 E 2c 2
1
12c
1 2 Y2
3 E 2c 2
2
Y bc 1
(a)
2
3
MY 1
2
Cubic equation for
Y
200 109 Pa
E
420 N m
M
2.4 m
b
20 mm
c
1
h
2
8 mm
Y
1
750 10
21
2
Y
350
Y
1
750 10
21
2
Y
273.44 106
(1.28 10 6 )
Solving by trial,
(b)
yY
Y
E
(292 106 )(2.4)
200 109
Y
0.020 m
0.008 m
292 106 Pa
3.504 10 3 m
Y
292 MPa
2 yY
7.01 mm
3.504 mm
thickness of elastic core
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545
PROBLEM 4.96
" (MPa)
300
The prismatic bar AB is made of an aluminum
alloy for which the tensile stress-strain diagram
is as shown. Assuming that the - diagram is
the same in compression as in tension,
determine (a) the radius of curvature of the bar
when the maximum stress is 250 MPa, (b) the
corresponding value of the bending moment.
versus y and use an
(Hint: For part b, plot
approximate method of integration.)
B
40 mm
M'
200
M
60 mm
A
100
0
0.005
0.010
#
SOLUTION
(a)
250 MPa
m
0.0064 from curve
c
b
1
h 30 mm 0.030 m
2
40 mm 0.040 m
1
m
c
(b)
250 106 Pa
m
0.0064
0.030
0.21333 m
1
Strain distribution:
4.69 m
y
c
m
Bending couple:
c
M
where the integral J is given by
c
1
u|
0
where
mu
y b dy
2b
c
0
u
y
1
2bc 2 0 u | | du
y | | dy
2bc 2 J
| du
Evaluate J using a method of numerial integration. If Simpson’s rule is used, the integration formula is
J
u
wu | |
3
where w is a weighting factor.
Using
u
0
0.25
0.5
0.75
1.00
u
0.25, we get the values given in the table below:
| |
0
0.0016
0.0032
0.0048
0.0064
J
M
| |, (MPa)
0
110
180
225
250
(0.25)(1215)
3
u | |, (MPa)
0
27.5
90
168.75
250
101.25 MPa
w
1
4
2
4
1
wu | |, (MPa)
0
110
180
675
250
1215
wu | |
101.25 106 Pa
(2)(0.040)(0.030)2 (101.25 106 )
7.29 103 N m
M
7.29 kN m
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546
PROBLEM 4.97
0.8 in.
B
M
" (ksi)
1.2 in.
A
50
The prismatic bar AB is made of a bronze alloy for which the tensile
stress-strain diagram is as shown. Assuming that the - diagram is
the same in compression as in tension, determine (a) the maximum
stress in the bar when the radius of curvature of the bar is 100 in.,
(b) the corresponding value of the bending moment. (See hint given
in Prob. 4.96.)
40
30
20
10
0
0.004
0.008
#
SOLUTION
(a)
100 in., b
c
0.6
100
m
(b)
0.8 in., c
0.6 in.
0.006
From the curve,
Strain distribution:
Bending couple:
m
c
M
where the integral J is given by
y
c
c
1
u
0
mu
y b dy
where u
2b
c
0
y | | dy
m
43.0 ksi
y
1
2bc 2 0 u | | du
2bc 2 J
| | du
Evaluate J using a method of numerial integration. If Simpson’s rule is used, the integration formula is
J
u
wu| |
3
where w is a weighting factor.
Using
u
0.25, we get the values given the table below:
u
| |
0
0.25
0.5
0.75
1.00
0
0.0015
0.003
0.0045
0.006
| |, ksi
0
25
36
40
43
u | |, ksi
0
6.25
18
30
43
J
M
w
1
4
2
4
1
(0.25)(224)
18.67 ksi
3
(2)(0.8)(0.6) 2 (18.67)
wu | |, ksi
0
25
36
120
43
224
wu | |
M
10.75 kip in.
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547
PROBLEM 4.98
"
A prismatic bar of rectangular cross section is made of an alloy for
which the stress-strain diagram can be represented by the relation
k n for
0 and
k n for
0 . If a couple M is
applied to the bar, show that the maximum stress is
#
M
1
m
2n Mc
I
3n
SOLUTION
Strain distribution:
y
c
m
where
mu
y
c
u
Bending couple:
c
M
c
1
0
2 bc 2
For
K
n
c
0
2b
y b dy
2bc 2
y | | dy
,
K
m
m
u
m
m
M
2bc 2 0 u
1
2bc
Recall that
Then
| |
2
m
mu
u
2
1
n
2 1n
m
2n 1 M
2 bc 2
I
c
1 b(2c)3
12 c
m
y
dy
| |
c
c
u | | du
n
Then
c
0
1
n
du
1
0
2 2
bc
3
2bc 2
mu
1
n
1 1 1n
du
m 0u
2n
bc 2
2n 1
1
bc 2
m
2 c
3 I
2n 1 Mc
3n
I
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548
PROBLEM 4.99
30 mm
Knowing that the magnitude of the horizontal force P is 8 kN,
determine the stress at (a) point A, (b) point B.
B
24 mm
A
D
P
45 mm
15 mm
SOLUTION
A
(30)(24)
720 mm 2
e
45
33 mm
I
c
M
(a)
A
P
A
Mc
I
12
720 10 6 m 2
0.033 m
1 3
1
bh
(30)(24)3 34.56 103 mm 4 34.56 10 9 m 4
12
12
1
(24 mm) 12 mm 0.012 m
P 8 103 N
2
Pe
(8 103 )(0.033)
8 103
720 10
6
264 N m
(264)(0.012)
34.56 10 9
102.8 106 Pa
102.8 MPa
A
(b)
B
P
A
Mc
I
8 103
720 10
6
(264)(0.012)
34.56 10 9
80.6 106 Pa
B
80.6 MPa
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549
PROBLEM 4.100
y
b
3 in.
6 kips
A short wooden post supports a 6-kip axial load as shown. Determine the
stress at point A when (a) b 0, (b) b 1.5 in., (c) b 3 in.
C
A
z
x
SOLUTION
A
I
S
P
(a)
b
0
M
P
A
(b)
b
(3) 2
28.27 in 2
r4
(3) 4 63.62 in 4
4
4
I
63.62
21.206 in 3
c
3
6 kips
M Pb
0
6
28.27
1.5 in. M
P
A
r2
M
S
0.212 ksi
212 psi
(6)(1.5)
9 kip in.
6
28.27
9
21.206
0.637 ksi
637 psi
(c)
b
3 in.
P
A
M
M
S
(6)(3)
18 kip in.
6
28.27
18
21.206
1.061 ksi
1061 psi
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550
P
r
r
P
PROBLEM 4.101
Two forces P can be applied separately or at the same time to a plate that is welded
to a solid circular bar of radius r. Determine the largest compressive stress in the
circular bar, (a) when both forces are applied, (b) when only one of the forces is
applied.
SOLUTION
For a solid section,
A
Both forces applied.
4
r 4, c
r
F Mc
A
I
F
4M
2
r
r3
Compressive stress
(a)
r 2, I
F
2 P, M
0
2P
r2
(b)
One force applied.
F
P, M
F
r2
Pr
4Pr
r2
5P
r2
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551
PROBLEM 4.102
y
30 mm
60 mm
A short 120 × 180-mm column supports the three axial loads
shown. Knowing that section ABD is sufficiently far from the
loads to remain plane, determine the stress at (a) corner A,
(b) corner B.
30 kN
20 kN
100 kN
C
z
x
A
D
90 mm
90 mm
B
120 mm
SOLUTION
A
S
M
(a)
A
(0.120 m)(0.180 m) 21.6 10 3 m 2
1
(0.120 m)(0.180 m)2 6.48 10 4 m 2
6
(30 kN)(0.03 m) (100 kN)(0.06 m)
5.10 kN m
P
A
M
S
150 103 N
21.6 10 3 m 2
5.10 103 N m
6.48 10 4 m3
A
(b)
B
P
A
M
S
3
150 10 N
21.6 10 3 m 2
0.926 MPa
3
5.10 10 N m
6.48 10 4 m3
B
14.81 MPa
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552
PROBLEM 4.103
80 mm
80 mm
P
C
1
As many as three axial loads, each of magnitude P 50 kN, can be applied to
the end of a W200 × 31.1 rolled-steel shape. Determine the stress at point A
(a) for the loading shown, (b) if loads are applied at points 1 and 2 only.
P
P
2
3
A
SOLUTION
For W200
(a)
31.3 rolled-steel shape.
Centric load:
A
3970 mm 2
c
1
d
2
I
31.3 106 mm 4
3P
50
1
(210)
2
50
3P
A
(b)
Ececentric loading:
e
50
50
M
Pe
(50 103 )(0.080)
2P
A
Mc
I
80 mm
50
3
m2
105 mm
0.105 m
31.3 10
150 kN
150 103
3.970 10 3
6
m4
150 103 N
37.783 106 Pa
37.8 MPa
0.080 m
100 103 N
2P
A
100 kN
4.000 10
4.0 103 N m
100 103
3.970 10 3
(4.0 103 )(0.105)
31.3 10 6
38.607 106 Pa
38.6 MPa
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553
PROBLEM 4.104
y
10 mm
10 mm
Two 10-kN forces are applied to a 20 × 60-mm rectangular bar as
shown. Determine the stress at point A when (a) b 0, (b) b 15 mm,
(c) b 25 mm.
A
30 mm
z
30 mm
10 kN
C
b
10 kN
x
25 mm
SOLUTION
A
1.2 10
3
m2
b
1
(0.020 m)(0.060 m)2 12 10 6 m3
6
0, M (10 kN)(0.025 m) 250 N m
A
(20 kN)/(1.2 10
S
(a)
(0.060 m)(0.020 m)
16.667 MPa
3
m2 )
(250 N m)/(12 10
6
m3 )
20.833 MPa
4.17 MPa
A
(b)
b
15 mm, M
A
(20 kN)/(1.2 10
16.667 MPa
(10 kN)(0.025 m
3
m2 )
0.015 m)
100 N m
(100 N m)/(12 10
6
m3 )
8.333 MPa
A
(c)
b
25 mm, M
0:
A
(20 kN)/(1.2 10
3
8.33 MPa
m2 )
A
16.67 MPa
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554
P
P'
P
PROBLEM 4.105
P'
Portions of a 12 12 -in. square bar have been bent
to form the two machine components shown.
Knowing that the allowable stress is 15 ksi,
determine the maximum load that can be applied
to each component.
1 in.
(a)
(b)
SOLUTION
The maximum stress occurs at point B.
B
B
where
ec
I
15 103 psi
15 ksi
P
A
K
1
A
A
(0.5)(0.5)
I
1
(0.5)(0.5)3
12
Mc
I
P
A
Pec
I
KP
1.0 in.
e
0.25 in 2
5.2083 10 3 in 4 for all centroidal axes.
(a)
(a)
(b)
c
(b)
0.25 in.
K
1
0.25
(1.0)(0.25)
5.2083 10 3
P
B
( 15 103 )
52
c
K
0.5
2
52 in
2
P
288 lb
P
209 lb
0.35355 in.
K
1
0.25
(1.0)(0.35355)
5.2083 10 3
P
B
( 15 103 )
71.882
K
71.882 in
2
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555
PROBLEM 4.106
P
Knowing that the allowable stress in section ABD is 80 MPa,
determine the largest force P that can be applied to the bracket shown.
A
D
B
18 mm
40 mm
12 mm
12 mm
SOLUTION
A
(24)(18)
432 mm 2
432 10
1
(24)(18)3 11.664 103 mm 4
12
1
(18) 9 mm 0.009 m
c
2
1
(18) 40 49 mm 0.049 m
e
2
On line BD, both axial and bending stresses are compressive.
I
Therefore
Solving for P gives
max
P
P
P
A
Mc
I
P
A
Pec
I
6
m2
11.664 10
9
m4
max
1
A
ec
I
80 106 Pa
1
432 10 6 m 2
(0.049 m)(0.009 m)
11.664 10 9 m 4
P
1.994 kN
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556
PROBLEM 4.107
P'
a
d
a
P
A milling operation was used to remove a portion of a solid bar
of square cross section. Knowing that a 30 mm, d 20 mm,
and all 60 MPa, determine the magnitude P of the largest
forces that can be safely applied at the centers of the ends of the
bar.
SOLUTION
A
e
Data:
ad ,
I
1
ad 3 ,
12
a d
2 2
P Mc
P
6 Ped
A
I
ad
ad 3
3P (a d )
P
KP
ad
ad 2
a
30 mm
K
1
(0.030)(0.020)
P
c
K
0.030 m
d
1
d
2
K
20 mm
0.020 m
(3)(0.010)
(0.030)(0.020)2
60 106
4.1667 103
1
ad
where
14.40 103 N
3(a d )
ad 2
4.1667 103 m
2
P
14.40 kN
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557
PROBLEM 4.108
P'
A milling operation was used to remove a portion of a solid bar of
square cross section. Forces of magnitude P 18 kN are applied at
the centers of the ends of the bar. Knowing that a 30 mm and
135 MPa, determine the smallest allowable depth d of the
all
milled portion of the bar.
a
d
P
a
SOLUTION
1
ad 3 ,
12
A
ad ,
e
a
2
d
2
P
A
Mc
I
3P
d2
I
2P
ad
Solving for d,
d
1
2
Data:
a
0.030 m,
d
1
(2)(135 106 )
16.04 10
3
P
ad
2P
a
c
1
d
2
Pec
I
P
ad
d2
or
2
18 103 N,
(2)(18 103 )
0.030
3P
P
ad
3P(a d )
ad 2
0
2P
a
12 P
P
2P
d
a
P 1 (a d ) 1 d
2
2
1
3
ad
12
135 106 Pa
2
12(18 103 )(135 106 )
m
(2)(18 103 )
0.030
d
16.04 mm
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558
12 kips
5 in.
PROBLEM 4.109
P
The two forces shown are applied to a rigid plate supported by a steel pipe of 8-in.
outer diameter and 7-in. inner diameter. Determine the value of P for which the
maximum compressive stress in the pipe is 15 ksi.
SOLUTION
15 ksi
all
I NA
A
4
(4 in.)4
4
(4 in.) 2
(3.5 in.)4
(3.5 in.)2
83.2 in 4
11.78 in 2
Max. compressive stress is at point B.
B
Q
A
Mc
I
15 ksi
1.019
13.981
0.325P
12 P
11.78 in 2
0.085P
(5P)(4.0 in.)
83.2 in 4
0.240 P
P
43.0 kips
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559
PROBLEM 4.110
d
P'
P
h
P'
P
An offset h must be introduced into a solid circular rod of
diameter d. Knowing that the maximum stress after the offset
is introduced must not exceed 5 times the stress in the rod
when it is straight, determine the largest offset that can be
used.
d
SOLUTION
For centric loading,
c
P
A
For eccentric loading,
e
P
A
Given
e
P
A
5
Phc
I
c
Phc
I
5
P
A
Phc
I
P
4
A
h
4I
cA
(4)
d
2
64
4
d4
d
2
1
d
2
h
0.500 d
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560
PROBLEM 4.111
d
P'
P
An offset h must be introduced into a metal tube of 0.75-in. outer
diameter and 0.08-in. wall thickness. Knowing that the maximum
stress after the offset is introduced must not exceed 4 times the
stress in the tube when it is straight, determine the largest offset
that can be used.
h
P'
P
d
SOLUTION
c
c1
1
d 0.375 in.
2
c t 0.375 0.08
c2
A
c12
0.295 in.
(0.3752
0.2952 )
0.168389 in 2
I
4
c4
c14
4
(0.3754
0.2954 )
9.5835 10 3 in 4
For centric loading,
cen
P
A
For eccentric loading,
ecc
P
A
ecc
4
hc
I
3
A
Phc
I
or
cen
h
P
A
3I
Ac
Phc
I
4
P
A
(3)(9.5835 10 3 )
(0.168389)(0.375)
h
0.455 in.
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561
PROBLEM 4.112
16 kips
2
4
1
A short column is made by nailing four 1 × 4-in. planks to a 4 × 4-in. timber. Using
an allowable stress of 600 psi, determine the largest compressive load P that can be
applied at the center of the top section of the timber column as shown if (a) the
column is as described, (b) plank 1 is removed, (c) planks 1 and 2 are removed,
(d) planks 1, 2, and 3 are removed, (e) all planks are removed.
3
SOLUTION
(a)
Centric loading:
M
P
A
0
A
(4
4)
32 in 2
4(1)(4)
P
A
(0.600 ksi)(32 in 2 )
P
19.20 kips
(b)
Eccentric loading:
M
P
A
Pe
A
(4)(4)
y
Ay
A
I
(I
(c)
Centric loading:
28 in 2
(3)(1)(3)
(1)(4)(2.5)
28
e
y
0.35714 in.
Ad 2 )
1
(6)(4)3
12
P
Pec
I
1
A
1
(4)(1)3
12
(6)(4)(0.35714)2
ec
I
P
(4)(1)(2.14286)2
(0.600 ksi)
(0.35714)(2.35714)
53.762
1
28
53.762 in 4
11.68 kips
P
A
M
0
A
(6)(4)
P
(0.600 ksi)(24)
24 in 2
14.40 kips
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562
PROBLEM 4.112 (Continued)
(d)
Eccentric loading:
M
Pe
A
(4)(4)
x
2.5
P
Centric loading:
(1)(4)(1)
2
1
(4)(5)3
12
I
(e)
P
A
M
Pec
I
20 in 2
x
0.5 in.
41.667 in 4
(0.600 ksi)
1
(0.5)(2.5)
20
41.667
0
e
7.50 kips
P
A
16 in 2
A
(4)(4)
P
(0.600 ksi)(16.0 in 2 )
9.60 kips
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563
1 in.
a
1.5 in.
PROBLEM 4.113
3 in.
a
0.75 in.
1.5 in.
A
A vertical rod is attached at point A to the cast
iron hanger shown. Knowing that the allowable
stresses in the hanger are all
5 ksi and
12 ksi , determine the largest downward
all
force and the largest upward force that can be
exerted by the rod.
3 in.
0.75 in.
B
Section a–a
SOLUTION
X
Ay
A
X
12.75 in 3
7.5 in 2
A
7.5 in 2
all
5 ksi
(1 3)(0.5)
(1 3)
all
1
(3)(1)3
12
0.25)(2.5)
0.75)
1.700 in.
1 3
bh
12
Ic
2(3
2(3
12 ksi
Ad 2
(3 1)(1.70
0.5)2
1
(1.5)(3)3
12
(1.5
3)(2.5
1.70)2
10.825 in 4
Ic
Downward force.
M
At D:
D
5 ksi
5
At E:
E
12 ksi
12
P(1.5 in.+1.70 in.)
P
A
Mc
I
P
(3.20) P(1.70)
7.5
10.825
P( 0.6359)
P
A
(3.20 in.) P
P
7.86 kips
P
21.95 kips
P
7.86 kips
Mc
I
P
(3.20) P(2.30)
7.5
10.825
P( 0.5466)
We choose the smaller value.
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564
PROBLEM 4.113 (Continued)
Upward force.
M
At D:
D
12 ksi
12
At E:
E
5 ksi
5
P(1.5 in.
P
A
1.70 in.)
Mc
I
P
(3.20) P(1.70)
7.5
10.825
P( 0.6359)
P
A
(3.20 in.) P
P
18.87 kips
Mc
I
P
(3.20) P(2.30)
7.5
10.825
P( 0.5466)
We choose the smaller value.
P
9.15 kips
P
9.15 kips
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565
PROBLEM 4.114
1 in.
a
1.5 in.
3 in.
a
1.5 in.
A
Solve Prob. 4.113, assuming that the vertical
rod is attached at point B instead of point A.
0.75 in.
3 in.
PROBLEM 4.113 A vertical rod is attached at
point A to the cast iron hanger shown. Knowing
that the allowable stresses in the hanger are
5 ksi and all
12 ksi, determine the
all
largest downward force and the largest upward
force that can be exerted by the rod.
0.75 in.
B
Section a–a
SOLUTION
X
Ay
A
X
12.75 in 3
7.5 in 2
A
7.5 in 2
all
5 ksi
(1 3)(0.5)
(1 3)
all
1
(3)(1)3
12
Ic
0.25)(2.5)
0.75)
1.700 in.
1 3
bh
12
Ic
2(3
2(3
12 ksi
Ad 2
(3 1)(1.70
0.5)2
1
(1.5)(3)3
12
(1.5
3)(2.5
1.70)2
10.825 in 4
Downward force.
all
M
At D:
D
12 ksi
12
At E:
E
5 ksi
5
5 ksi
(2.30 in.
P
A
12 ksi
all
1.5 in.)
(3.80 in.) P
Mc
I
P
(3.80) P(1.70)
7.5
10.825
P( 0.4634)
P
A
P
25.9 kips
P
5.32 kips
P
5.32 kips
Mc
I
P
(3.80) P(2.30)
7.5
10.825
P( 0.9407)
We choose the smaller value.
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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566
PROBLEM 4.114 (Continued)
Upward force.
all
M
At D:
D
5 ksi
5
At E:
E
12 ksi
12
5 ksi
(2.30 in.
P
A
12 ksi
all
1.5 in.)P
Mc
I
P
(3.80) P(1.70)
7.5
10.825
P( 0.4634)
P
A
(3.80 in.)P
P
10.79 kips
P
12.76 kips
P
10.79 kips
Mc
I
P
(3.80) P(2.30)
7.5
10.825
P( 0.9407)
We choose the smaller value.
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
567
PROBLEM 4.115
2 mm radius
A
P
Knowing that the clamp shown has been tightened until
P 400 N, determine (a) the stress at point A,
(b) the stress at point B, (c) the location of the neutral
axis of section a a.
P'
32 mm
20 mm
a
B
a
4 mm
Section a–a
SOLUTION
Cross section:
Rectangle
Circle
(20 mm)(4 mm)
y1
1
(20 mm)
2
10 mm
A2
(2 mm)2
4 mm 2
y2
y
cA
20
d1
11.086
d2
18
y
2
18 mm
(80)(10)
80
(4 )(18)
4
1.086 mm
11.086
6.914 mm
I1
A1d12
1
(4)(20)3
12
I2
I2
A2d 22
(2) 4
I
I1
I2
A
A1
A2
(80)(1.086) 2
(4 )(6.914) 2
3.374 103 mm 4
92.566 mm 2
11.086 mm
8.914 mm
10
I1
4
20
Ay
A
cB
80 mm 2
A1
2.761 103 mm 4
0.613 103 mm 4
3.374 10
92.566 10
6
9
m4
m2
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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568
PROBLEM 4.115 (Continued)
(a)
e
32
8.914
M
Pe
(400 N)(0.040914 m)
16.3656 N m
Mc
I
(16.3656)(8.914 10 3 )
3.374 10 9
P
A
4.321 106
400
92.566 10
6
43.23 106
47.55 106 Pa
A
47.6 MPa
Point B:
B
P
A
Mc
I
4.321 106
(c)
0.040914 m
Point A:
A
(b)
40.914 mm
Neutral axis:
400
92.566 10
6
53.72 106
(16.3656)(11.086)
3.374 10 9
49.45 106 Pa
B
49.4 MPa
By proportions,
a
47.55
a
20
47.55 49.45
9.80 mm
9.80 mm below top of section
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569
PROBLEM 4.116
P
a
a
The shape shown was formed by bending a thin steel plate. Assuming that the
thickness t is small compared to the length a of a side of the shape, determine the
stress (a) at A, (b) at B, (c) at C.
90$
t
B
C
A
P'
SOLUTION
Moment of inertia about centroid:
I
a
1
2 2t
12
3
2
1 3
ta
12
Area:
A
2 2t
a
2
2at , c
Pec
I
P
2at
P
A
P
A
(b)
Pec
I
P
2at
P
B
P
A
(c)
C
(a)
a
2 2
1
ta
12
a
2
a
2 2
a
2 2
3
A
P
2at
B
2P
at
C
P
2at
a
2
1
ta3
12
A
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570
PROBLEM 4.117
P
Three steel plates, each of 25 150-mm cross section, are welded
together to form a short H-shaped column. Later, for architectural
reasons, a 25-mm strip is removed from each side of one of the
flanges. Knowing that the load remains centric with respect to the
original cross section, and that the allowable stress is 100 MPa,
determine the largest force P (a) that could be applied to the original
column, (b) that can be applied to the modified column.
50 mm
50 mm
SOLUTION
(a)
P
A
Centric loading:
A
11.25 103 mm 2
(3)(150)(25)
P
11.25 10 3 m 2
( 100 106 )(11.25 10 3 )
A
1.125 106 N
(b)
P
1125 kN
Eccentric loading (reduced cross section):
A, 103 mm 2
y , mm
A y (103 mm3 )
3.75
87.5
328.125
76.5625
3.75
0
0
10.9375
2.50
87.5
218.75
98.4375
10.00
d , mm
109.375
Y
Ay
A
109.375 103
10.00 103
10.9375 mm
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
571
PROBLEM 4.117 (Continued)
The centroid lies 10.9375 mm from the midpoint of the web.
I1
I2
I3
1
b1h13
12
1
b2h23
12
1
b3h33
12
A2d 22
A3d32
I2
I3
I
I1
c
10.9375
M
Pe
P
A
K
P
75
K
54.012 106 mm 4
25
Mc
I
P
A
10.4375 mm
Pec
I
1
10.00 10
( 100 106 )
122.465
3
22.177 106 mm 4
7.480 106 mm 4
24.355 106 mm 4
54.012 10 6 m 4
110.9375 mm
e
where
ec
I
1
A
1
(150)(25)3 (3.75 103 )(76.5625)2
12
1
(25)(150)3 (3.75 103 )(10.9375)2
12
1
(100)(25)3 (2.50 103 )(98.4375) 2
12
A1d12
KP
0.1109375 m
10.4375 10 3 m
A
10.00 10 3 m 2
(101.9375 10 3 )(0.1109375)
54.012 10 6
817 103 N
122.465 m
2
P
817 kN
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
572
PROBLEM 4.118
y
P
y
B
3 in.
y
x
3 in.
B
A vertical force P of magnitude 20 kips is applied
at point C located on the axis of symmetry of the
cross section of a short column. Knowing that
y 5 in., determine (a) the stress at point A,
(b) the stress at point B, (c) the location of the
neutral axis.
2 in.
C
A
4 in.
A
2 in.
x
2 in.
1 in.
(a)
(b)
SOLUTION
Locate centroid.
A, in 2
12
8
20
Eccentricity of load: e 5
(a)
I1
1
(6)(2)3
12
I
I1
I2
Stress at A : c A
Ay , in 3
60
16
76
y , in.
5
2
Part
3.8
Stress at B : cB
(12)(1.2) 2
21.28 in 4
3.8 in.
1
(2)(4)3
12
I2
(8)(1.8)2
36.587 in 4
57.867 in 4
3.8 in.
P
A
6
3.8
Pec A
I
20
20
20(1.2)(3.8)
57.867
PecB
I
20
20
20(1.2)(2.2)
57.867
A
0.576 ksi
2.2 in.
P
A
B
(c)
76
20
1.2 in.
A
(b)
Ay
A
y
Location of neutral axis:
0
a
I
Ae
P Pea
0
A
I
57.867
2.411 in.
(20)(1.2)
Neutral axis lies 2.411 in. below centroid or 3.8
2.411
ea
I
B
1.912 ksi
1
A
1.389 in. above point A.
Answer: 1.389 in. from point A
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
573
PROBLEM 4.119
y
P
y
B
3 in.
y
x
3 in.
B
A vertical force P is applied at point C located
on the axis of symmetry of the cross section of
a short column. Determine the range of values
of y for which tensile stresses do not occur in
the column.
2 in.
C
A
4 in.
A
x
2 in.
2 in.
1 in.
(a)
(b)
SOLUTION
Locate centroid.
A, in 2
y , in.
Ay , in 3
5
2
60
16
76
12
8
20
Eccentricity of load:
e
y
I1
1
(6)(2)3
12
I
I1
If stress at A equals zero,
If stress at B equals zero,
y
3.8 in.
3.8 in.
A
P
A
e
I
Ac A
e
y
3.8 in.
3.8 in.
(12)(1.2) 2
cA
B
76
20
21.28 in 4
1
(2)(4)3
12
I2
(8)(1.8) 2
36.587 in 4
57.867 in 4
I2
cB
e
Ai yi
Ai
y
6
Pec A
I
3.8
ec A
I
0
57.867
(20)(3.8)
0.761 in.
y
0.761
3.8
4.561 in.
2.2 in.
P PecB
0
A
I
57.867
I
(20)(2.2)
AcB
1.315
1
A
3.8
ecB
I
1
A
1.315 in.
2.485 in.
Answer: 2.49 in.
y
4.56 in.
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
574
PROBLEM 4.120
P
P
The four bars shown have the same cross-sectional area. For the given
loadings, show that (a) the maximum compressive stresses are in the
ratio 4:5:7:9, (b) the maximum tensile stresses are in the ratio 2:3:5:3.
(Note: the cross section of the triangular bar is an equilateral triangle.)
P
P
SOLUTION
Stresses:
At A,
A
P
A
Pec A
I
P
1
A
At B,
B
P
A
PecB
I
P AecB
A
I
A1
1 4
a , c A cB
12
1
1
a
(a 2 ) a
P
2
2
1
1 2
A
a
12
a 2 , I1
A
B
A2
A
B
P
A
c2
(a 2 )
a2
1
1
a
a
2
2
1 2
a
12
c
a
Aec A
I
1
1
a, e
2
1
a
2
A
1
, I2
B
4
c4 , e
P
( c 2 )(c)(c)
1
A2
c4
4
P
A1
2
P
A1
5
P
A2
3
P
A2
c
A
P ( c 2 )(c)(c)
1
A2
4
c
4
4
B
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575
PROBLEM 4.120 (Continued)
A3
A
B
a2
2
a I3
2
c
(a 2 )
P
1
A3
P
A3
1 4
a
12
e
c
2
2
a
a
2
2
1 4
a
12
A
2
2
a
a
2
2
1 4
a
12
1
A4
1
(s)
2
3
s
2
3 2
s
4
I4
1
s
36
3
s
2
cA
2 3
s
3 2
(a 2 )
A
B
3
s
P
A4
cB
5
P
A3
3 2
s
4
s
3
9
P
A4
3
P
A4
s
s
3
A
3 4
s
96
3 2
s
4
P
A3
3 4
s
96
e
3
P
1
A4
B
7
s
3
3 4
s
96
s
2 3
1
B
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576
PROBLEM 4.121
25 mm
An eccentric force P is applied as shown to a steel bar of
25 90-mm cross section. The strains at A and B have been
measured and found to be
30 mm
A
90 mm
45 mm
B
350
A
P
d
70
B
Knowing that E 200 GPa, determine (a) the distance d, (b) the
magnitude of the force P.
15 mm
SOLUTION
h 15 45 30 90 mm
A bh (25)(90)
I
yA
1
h 45 mm
2
2.25 10 3 m 2
25 mm
b
2.25 103 mm 2
Subtracting,
A
A
E
A
(200 109 )(350 10 6 )
B
E
B
(200 109 )( 70 10 6 )
M
A( y A B yB
y A yB
Pd
d
MyA
I
(1)
B
P
A
MyB
I
(2)
M ( y A yB )
I
M
I( A
B)
y A yB
M
P
14 106 Pa
P
A
B
A)
70 106 Pa
A
(1.51875 10 6 )(84 106 )
0.045
Multiplying (2) by y A and (1) by yB and subtracting,
(a)
0.045 m
1 3 1
(25)(90)3 1.51875 106 mm 4 1.51875 10 6 m 4
bh
12
12
60 45 15 mm 0.015 m
30 mm
0.030 m
yB 15 45
Stresses from strain gages at A and B:
P
c
yA
B
yB
A
( yA
2835 N m
yB )
P
A
(2.25 10 3 )[(0.015)( 14 106 ) ( 0.030)(70 106 )]
0.045
2835
94.5 103
0.030 m
(b)
94.5 103 N
d
P
30.0 mm
94.5 kN
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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577
PROBLEM 4.122
25 mm
Solve Prob. 4.121, assuming that the measured strains are
A
90 mm
B
600
A
30 mm
45 mm
P
PROBLEM 4.121 An eccentric force P is applied as shown to a
steel bar of 25 90-mm cross section. The strains at A and B have
been measured and found to be
d
350
A
15 mm
420
B
B
70
Knowing that E 200 GPa, determine (a) the distance d, (b) the
magnitude of the force P.
SOLUTION
h 15 45 30 90 mm
A bh
b
1
h 45 mm
2
2.25 10 3 m 2
2.25 103 mm 2
(25)(90)
yA
Stresses from strain gages at A and B:
A
E
A
(200 109 )(600 10 6 ) 120 106 Pa
B
E
B
(200 109 )(420 10 6 ) 84 106 Pa
A
P
A
My A
I
(1)
B
P
A
MyB
I
(2)
Subtracting,
A
B
M
M ( y A yB )
I
I( A
B)
y A yB
Multiplying (2) by y A and (1) by yB and subtracting,
(a)
M
A( y A B yB
y A yB
Pd
0.045 m
1 3 1
(25)(90)3 1.51875 106 mm 4 1.51875 10 6 m 4
bh
12
12
60 45 15 mm 0.015 m yB 15 45
30 mm
0.030 m
I
P
c
25 mm
d
A)
(1.51875 10 6 )(36 106 )
0.045
yA
B
yB
A
( yA
yB )
(2.25 10 3 )[(0.015)(84 106 ) ( 0.030)(120 106 )]
0.045
M
1215
5 10 3 m
3
P
243 10
(b)
1215 N m
P
A
243 103 N
d
5.00 mm
P
243 kN
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
578
PROBLEM 4.123
P'
The C-shaped steel bar is used as a dynamometer to determine the magnitude P
of the forces shown. Knowing that the cross section of the bar is a square of side
40 mm and that strain on the inner edge was measured and found to be 450 ,
determine the magnitude P of the forces. Use E 200 GPa.
40 mm
80 mm
P
SOLUTION
At the strain gage location,
E
A
(200 109 )(450 10 6 )
1600 mm 2
(40)(40)
90 106 Pa
1600 10
e
1
(40)(40)3 213.33 103 mm 4
12
80 20 100 mm 0.100 m
c
20 mm
I
K
P
6
m2
213.33 10
9
m4
0.020 m
P
A
Mc
I
P
A
Pec
I
1
A
ec
I
K
90 106
10.00 103
KP
1
1600 10
6
(0.100)(0.020)
213.33 10 9
10.00 103 m
9.00 103 N
2
P
9.00 kN
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
579
P
y
6 in.
PROBLEM 4.124
6 in.
10 in.
Q
B
A
A short length of a rolled-steel column supports a rigid plate
on which two loads P and Q are applied as shown. The strains
at two points A and B on the centerline of the outer faces of the
flanges have been measured and found to be
x
x
z
A
z
A = 10.0 in2
Iz = 273 in4
6
in./in.
29
6
400 10
A
Knowing that E
each load.
B
300 10
6
in./in.
10 psi, determine the magnitude of
SOLUTION
Stresses at A and B from strain gages:
A
E
A
(29 106 )( 400 10 6 )
11.6 103 psi
B
E
B
(29 106 )( 300 10 6 )
8.7 103 psi
Centric force:
F
P Q
Bending couple:
M
6 P 6Q
c
F
A
A
5 in.
Mc
I
11.6 103
F
A
B
Mc
I
8.7 103
P Q
10.0
(6 P 6Q)(5)
273
0.00989 P 0.20989Q
P Q
10.0
(1)
(6 P 6Q )(5)
273
0.20989 P 0.00989Q
(2)
Solving (1) and (2) simultaneously,
P
44.2 103 lb
Q
57.3 103 lb 57.3 kips
44.2 kips
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
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580
PROBLEM 4.125
y
P
A single vertical force P is applied to a short steel post as shown.
Gages located at A, B, and C indicate the following strains:
z
x
500
A
C
A
1000
200
C
Knowing that E 29 106 psi, determine (a) the magnitude of
P, (b) the line of action of P, (c) the corresponding strain at the
1.5 in.
hidden edge of the post, where x
2.5 in. and z
B
3 in.
5 in.
B
SOLUTION
Ix
Mx
1
(5)(3)3 11.25 in 4
12
Pz
Mz
Px
2.5 in., xB
xA
zA
1.5 in., z B
2.5 in., xC
1.5 in., zC
1.5 in., z D
6
6
E
A
(29 10 )( 500 10 )
B
E
B
(29 106 )( 1000 10 6 )
A
31.25 in 4
2.5 in., xD
A
C
A
(5)(3)
15 in 2
2.5 in.
1.5 in.
14,500 psi
29, 000 psi
14.5 ksi
29 ksi
E C (29 106 )( 200 10 6 )
5800 psi
5.8 ksi
P M x z A M z xA
0.06667 P 0.13333M x 0.08M z
A
Ix
Iz
(1)
B
P
A
M x zB
Ix
M z xB
Iz
0.06667 P
0.13333M x
0.08M z
(2)
C
P
A
M x zC
Ix
M z xC
Iz
0.06667 P
0.13333M x
0.08M z
(3)
Substituting the values for
gives
Mx
87 kip in.
x
Mz
P
Mx
P
z
D
A
,
Mz
B
, and
C
into (1), (2), and (3) and solving the simultaneous equations
(a) P
90.625 kip in.
90.625
152.25
87
152.25
P
A
M x zD
Ix
M z xD
Iz
(0.06667)(152.25)
(c)
1
(3)(5)3
12
Iz
Strain at hidden edge:
0.06667 P
(0.13333)(87)
D
E
0.13333M x
(b) x
0.595 in.
z
0.571 in.
0.08M z
(0.08)( 90.625)
8.70 103
29 106
152.3 kips
8.70 ksi
300
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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581
PROBLEM 4.126
b ! 40 mm
A
a ! 25 mm
d
B
D
P
The eccentric axial force P acts at point D, which must be located
25 mm below the top surface of the steel bar shown. For P 60 kN,
determine (a) the depth d of the bar for which the tensile stress at
point A is maximum, (b) the corresponding stress at point A.
C
20 mm
SOLUTION
1
bd 3
12
1
1
d
e
d a
2
2
P Pec
A
I
1
1
P 1 12 2 d a 2 d
b d
d3
A bd
c
A
A
(a)
Depth d for maximum
d A
dd
(b)
A
I
A:
P
b
60 103
4
3
40 10
75 10
6a
d2
Differentiate with respect to d.
4
d2
3
P 4
b d
12a
d3
d
0
(6)(25 10 3 )
(75 10 3 )2
40 106 Pa
3a
d
A
75 mm
40 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
582
y
M ! 300 N · m
A
z
PROBLEM 4.127
" ! 60#
B
16 mm
C
16 mm
The couple M is applied to a beam of the cross section shown in a plane
forming an angle with the vertical. Determine the stress at (a) point A,
(b) point B, (c) point D.
D
40 mm
40 mm
SOLUTION
1
(80)(32)3 218.45 103 mm 4 218.45 10 9 m 4
12
1
(32)(80)3 1.36533 106 mm 4 1.36533 10 6 m 4
12
yB
yD 16 mm
Iz
Iy
yA
zA
My
(a)
A
300cos 30
M z yA
Iz
zB
zD
40 mm
259.81 N m
M y zA
Iy
Mz
(150)(16 10 3 )
218.45 10 9
300sin 30
150 N m
(259.81)(40 10 3 )
1.36533 10 6
3.37 106 Pa
(b)
B
M z yB
Iz
M y zB
Iy
(150)(16 10 3 )
218.45 10 9
(259.81)( 40 10 3 )
1.36533 10 6
18.60 106 Pa
(c)
D
M z yD
Iz
M y zD
Iy
(150)( 16 10 3 )
218.45 10 9
3.37 MPa
A
18.60 MPa
B
(259.81)( 40 10 3 )
1.36533 10 6
3.37 106 Pa
D
3.37 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
583
PROBLEM 4.128
y
! " 30#
A
The couple M is applied to a beam of the cross section shown in a plane
forming an angle with the vertical. Determine the stress at (a) point A,
(b) point B, (c) point D.
B
M " 400 lb · m
0.6 in.
z
C
0.6 in.
D
0.4 in.
SOLUTION
Iz
Iy
yA
zA
My
(a)
A
400 cos 60
200 lb in.,
M y zA
M z yA
Iz
Iy
1
(0.4)(1.2)3 57.6 10 3 in 4
12
1
(1.2)(0.4)3 6.40 10 3 in 4
12
yB
yD 0.6 in.
zB
Mz
1
(0.4) 0.2 in.
2
zD
400 sin 60
( 346.41)(0.6)
57.6 10 3
346.41 lb in.
(200)(0.2)
6.40 10 3
9.86 103 psi 9.86 ksi
(b)
B
M y zB
M z yB
Iz
Iy
( 346.41)(0.6)
57.6 10 6
2.64 103 psi
(c)
D
M z yD
Iz
M y zD
Iy
( 346.41)( 0.6)
57.6 10 3
9.86 103 psi
(200)( 0.2)
6.4 10 3
2.64 ksi
(200)( 0.2)
6.40 10 3
9.86 ksi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
584
y
A
M ! 25 kN · m
PROBLEM 4.129
" ! 15#
The couple M is applied to a beam of the cross section shown in a
with the vertical. Determine the stress at
plane forming an angle
(a) point A, (b) point B, (c) point D.
B
80 mm
C
z
20 mm
80 mm
D
30 mm
SOLUTION
My
25sin 15
6.4705 kN m
Mz
25cos15
24.148 kN m
1
1
(80)(90)3
(80)(30)3
12
12
5.04 10 6 m 4
Iy
Iy
Iz
1
(90)(60)3
3
M yz
Stress:
(a)
A
1
(60)(20)3
3
Iy
5.04 106 mm 4
1
(30)(100)3 16.64 106 mm 4
3
16.64 10 6 m 4
Mzy
Iz
(6.4705 kN m)(0.045 m)
5.04 10 6 m 4
(24.148 kN m)(0.060 m)
16.64 10 6 m 4
57.772 MPa 87.072 MPa
(b)
B
(6.4705 kN m)( 0.045 m)
5.04 10 6 m 4
A
(24.148 kN m)(0.060 m)
16.64 10 6 m 4
57.772 MPa 87.072 MPa
(c)
D
(6.4705 kN m)( 0.015 m)
5.04 10 6 m 4
29.3 MPa
B
144.8 MPa
D
125.9 MPa
(24.148 kN m)( 0.100 m)
16.64 10 6 m 4
19.257 MPa 145.12 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
585
y
PROBLEM 4.130
" ! 20#
M ! 10 kip · in.
A
z
2 in.
C
B
D
2 in.
3 in.
The couple M is applied to a beam of the cross section shown in a plane
forming an angle with the vertical. Determine the stress at (a) point A,
(b) point B, (c) point D.
3 in.
4 in.
SOLUTION
Locate centroid.
A, in 2
z , in.
Az , in 3
16
1
16
8
2
16
Σ
24
0
The centroid lies at point C.
yA
1
1
(2)(8)3
(4)(2)3 88 in 4
12
12
1
1
(8)(2)3
(2)(4)3 64 in 4
3
3
4 in.
yB 1 in.,
yD
zA
zB
Iz
Iy
(a)
M z yA
Iz
M y zA
A
(b)
M z yB
Iz
M y zB
B
(c)
M z yD
Iz
M y zD
D
Iy
Iy
Iy
zD
4 in.,
0
Mz
10 cos 20
9.3969 kip in.
My
10 sin 20
3.4202 kip in.
(9.3969)(1)
88
(3.4202)( 4)
64
(9.3969)( 1)
88
(3.4202)( 4)
64
(9.3969)( 4)
88
(3.4202)(0)
64
A
0.321 ksi
0.107 ksi
B
D
0.427 ksi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
586
PROBLEM 4.131
y
M 5 60 kip · in.
b 5 508
A
The couple M is applied to a beam of the cross section shown in a
plane forming an angle with the vertical. Determine the stress at
(a) point A, (b) point B, (c) point D.
B
3 in.
z
C
3 in.
D
1 in.
2.5 in. 2.5 in.
5 in.
5 in.
1 in.
SOLUTION
Mz
My
yA
zA
(a)
A
M z yA
Iz
60 sin 40
60 cos 40
yB
yD
zB
zD
38.567 kip in.
45.963 kip in.
3 in.
5 in.
Iz
1
(10)(6)3
12
2
Iy
1
(6)(10)3
12
2
M y zA
Iy
4
4
( 38.567)(3)
178.429
(1) 2
178.429 in 4
(1) 4
(1) 2 (2.5) 2
459.16 in 4
(45.963)(5)
459.16
1.149 ksi
(b)
B
M z yB
Iz
M y zB
Iy
( 38.567)(3)
178.429
A
(45.963)( 5)
459.16
0.1479
(c)
D
M z yD
Iz
M y zD
Iy
( 38.567)( 3)
178.429
1.149 ksi
B
0.1479 ksi
D
1.149 ksi
(45.963)( 5)
459.16
1.149 ksi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
587
PROBLEM 4.132
y
M 5 75 kip · in.
The couple M is applied to a beam of the cross section shown in a
plane forming an angle with the vertical. Determine the stress at
(a) point A, (b) point B, (c) point D.
b 5 758
A
2.4 in.
B
1.6 in. z
C
D
4 in.
4.8 in.
SOLUTION
Iz
Iy
yA
zA
(a)
M z yA
Iz
M y zA
A
(b)
M z yB
Iz
M y zB
B
(c)
M z yD
Iz
M y zD
D
Iy
Iy
Iy
1
1
(4.8)(2.4)3
(4)(1.6)3 4.1643 in 4
12
12
1
1
(2.4)(4.8)3
(1.6)(4)3 13.5851 in 4
12
12
yB
yD 1.2 in.
zB
2.4 in.
zD
Mz
75sin15
19.4114 kip in.
My
75cos15
72.444 kip in.
(19.4114)(1.2)
4.1643
(72.444)(2.4)
13.5851
(19.4114)(1.2)
4.1643
(72.444)( 2.4)
13.5851
(19.4114)( 1.2)
4.1643
(72.444)( 2.4)
13.5851
A
B
D
7.20 ksi
18.39 ksi
7.20 ksi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
588
PROBLEM 4.133
y
! " 30#
M " 100 N · m
The couple M is applied to a beam of the cross section shown in a
plane forming an angle with the vertical. Determine the stress at
(a) point A, (b) point B, (c) point D.
B
z
C
D
A
r " 20 mm
SOLUTION
Iz
8
r4
2
4r
3
r2
(0.109757)(20) 4
Iy
yA
yB
A
8
8
9
17.5611 10
r4
3
mm 4
17.5611 10
(20) 4
62.832 10 3 mm 4 62.832 10
8
8
4r
(4)(20)
yD
8.4883 mm
3
3
20 8.4883 11.5117 mm
r4
zA
(a)
2
20 mm
zD
zB
Mz
100 cos30
86.603 N m
My
100sin 30
50 N m
M z yA
Iz
M y zA
Iy
9
9
m4
m4
0
(86.603)( 8.4883 10 3 )
17.5611 10 9
(50)(20 10 3 )
62.832 10 9
57.8 106 Pa
(b)
B
M z yB
Iz
M y zB
Iy
A
3
(86.603)(11.5117 10 )
17.5611 10 9
(50)(0)
62.832 10
9
56.8 106 Pa
(c)
D
M z yD
Iz
M y zD
Iy
56.8 MPa
B
3
(86.603)( 8.4883 10 )
17.5611 10 9
25.9 106 Pa
57.8 MPa
3
(50)( 20 10 )
62.832 10 9
D
25.9 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
589
W310 $ 38.7 15#
PROBLEM 4.134
B
A
C
310 mm
The couple M is applied to a beam of the cross section shown in a plane
with the vertical. Determine the stress at
forming an angle
(a) point A, (b) point B, (c) point D.
M " 16 kN · m
D
E
165 mm
SOLUTION
For W310 38.7 rolled steel shape,
(a)
tan
Iz
tan
I
84.9 10
7.20 10
Iz
84.9 106 mm 4
84.9 10 6 m 4
Iy
7.20 106 mm 4
7.20 10 6 m 4
yA
yB
1
yD
2
zA
zE
zB
yE
zD
1
(310) 155 mm
2
1
(165) 82.5 mm
2
Mz
(16 103 ) cos 15
15.455 103 N m
My
(16 103 ) sin 15
4.1411 103 N m
6
6
tan 15
3.1596
72.4
72.4 15 57.4
(b)
Maximum tensile stress occurs at point E.
E
M z yE
Iz
M y zE
Iy
(15.455 103 )( 155 10 3 )
84.9 10 6
75.7 106 Pa
(4.1411 103 )(82.5 10 3 )
7.20 10 6
75.7 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
590
PROBLEM 4.135
208
S6 3 12.5
A
M 5 15 kip · in.
The couple M acts in a vertical plane and is applied to a beam oriented
as shown. Determine (a) the angle that the neutral axis forms with the
horizontal, (b) the maximum tensile stress in the beam.
B
C
E
3.33 in.
6 in.
D
SOLUTION
For S6 12.5 rolled steel shape,
Iz
22.0 in 4
Iy
1.80 in 4
zE
yA
(a)
tan
Iz
tan
Iy
zA
zB
yB
yD
zD
yE
1
(3.33) 1.665 in.
2
1
(6) 3 in.
2
Mz
15 sin 20
5.1303 kip in.
My
15 cos 20
14.095 kip in.
22.0
tan (90
1.80
20 ) 33.58
88.29
88.29
(b)
70
18.29
Maximum tensile stress occurs at point D.
D
M z yD
Iz
M y zD
Iy
(5.1303)( 3)
22.0
(14.095)(1.665)
13.74 ksi
1.80
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
591
PROBLEM 4.136
5#
C150 ! 12.2
B
A
M " 6 kN · m
C
152 mm
The couple M acts in a vertical plane and is applied to a beam oriented as
shown. Determine (a) the angle that the neutral axis forms with the
horizontal, (b) the maximum tensile stress in the beam.
13 mm
E
D
48.8 mm
SOLUTION
My
6 sin 5
0.52293 kN m
Mz
6 cos 5
5.9772 kN m
C150 12.2
(a)
Iy
0.286 10
Iz
5.45 10
6
6
m4
m4
Neutral axis:
tan
Iz
tan
Iy
tan
1.66718
5.45 10 6 m 4
tan 5
0.286 10 6 m 4
59.044
5
(b)
Maximum tensile stress at E: y
M y zE
E
Iy
Mz yE
Iz
E
54.0
76 mm, z E
13 mm
(0.52293 kN m)(0.013 m)
0.286 10 6 m 4
(5.9772 kN m)( 0.076 m)
5.45 10 6 m 4
E
107.1 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
592
y'
30"
PROBLEM 4.137
B
50 mm
A
5 mm
5 mm
C
M ! 400 N · m
z'
The couple M acts in a vertical plane and is applied to a beam oriented
as shown. Determine (a) the angle that the neutral axis forms with the
horizontal, (b) the maximum tensile stress in the beam.
D
E
18.57 mm
5 mm
50 mm
Iy' ! 281 # 103 mm4
Iz' ! 176.9 # 103 mm4
SOLUTION
Iz
176.9 103 mm 4
Iy
281 103 mm 4
yE
(a)
tan
Iz
tan
Iy
18.57 mm, z E
176.9 10
281 10
m4
m4
25 mm
Mz
400 cos 30
346.41 N m
My
400sin 30
200 N m
176.9 10 9
tan 30
281 10 9
9
9
0.36346
19.97
30
(b)
19.97
10.03
Maximum tensile stress occurs at point E.
E
M z yE
Iz
M y zE
Iy
(346.41)( 18.57 10 3 )
176.9 10 9
36.36 106
17.79 106
(200)(25 10 3 )
281 10 9
54.2 106 Pa
E
54.2 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
593
458
PROBLEM 4.138
y'
B
z'
0.859 in.
M 5 15 kip · in.
C
A
1
2
in.
4 in.
D
4 in.
The couple M acts in a vertical plane and is applied to a beam oriented
as shown. Determine (a) the angle that the neutral axis forms with the
horizontal, (b) the maximum tensile stress in the beam.
4 in.
Iy' 5 6.74 in4
Iz' 5 21.4 in4
SOLUTION
Iz
21.4 in 4
zA
zB
yA
Mz
(a)
0.859 in.
4 in.
My
Iy
yB
zD
4 in.
15 sin 45
15 cos 45
6.74 in 4
4 0.859 in.
yD
3.141 in.
0.25 in.
10.6066 kip in.
10.6066 kip in.
Angle of neutral axis:
tan
Iz
tan
Iy
21.4
tan ( 45 ) 3.1751
6.74
72.5
72.5
(b)
27.5
45
The maximum tensile stress occurs at point D.
D
M z yD
Iz
M y zD
Iy
0.12391 4.9429
(10.6066)( 0.25)
21.4
( 10.6066)( 3.141)
6.74
D
5.07 ksi
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Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
594
y'
20"
PROBLEM 4.139
B
10 mm
A
The couple M acts in a vertical plane and is applied to a beam
oriented as shown. Determine (a) the angle that the neutral
axis forms with the horizontal, (b) the maximum tensile stress
in the beam.
6 mm
M ! 120 N · m
C
z'
10 mm
D
6 mm
Iy' ! 14.77 # 103 mm4
Iz' ! 53.6 # 103 mm4
E
10 mm
10 mm
SOLUTION
(a)
Iz
53.6 103 mm 4
Iy
14.77 103 mm 4
14.77 10 9 m 4
Mz
120 sin 70
112.763 N m
My
120 cos 70
41.042 N m
Angle of neutral axis:
20
tan
(b)
53.6 10 9 m 4
Iz
tan
Iy
53.6 10 9
tan 20
14.77 10 9
52.871
52.871
20
1.32084
32.9
The maximum tensile stress occurs at point E.
yE
zE
E
16 mm
10 mm
M z yE
Iz
0.016 m
0.010 m
M y zE
Iy
(112.763)( 0.016)
53.6 10 9
61.448 106 Pa
(41.042)(0.010)
14.77 10 9
E
61.4 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
595
PROBLEM 4.140
208
A
90 mm
M 5 750 N · m
C
The couple M acts in a vertical plane and is applied to a beam oriented as
shown. Determine (a) the angle that the neutral axis forms with the horizontal,
(b) the maximum tensile stress in the beam.
B
30 mm
25 mm
25 mm
SOLUTION
(a)
My
750 sin 20
My
256.5 N m
Mz
750 cos 20
Mz
704.8 N m
1
(90)(25)3
12
0.2344 106 mm 4
Iy
2
Iy
0.2344 10 6 m 4
Iz
1
(50)(90)3 1.0125 106 mm 4
36
1.0125 10 6 m 4
Neutral axis:
tan
Iz
tan
Iy
tan
1.5724
20
1.0125 10 6 m 4
tan 20
0.2344 106 m 4
57.5
57.5
30
37.5
37.5
(b)
Maximum tensile stress, at D: y D
My zD
D
Iy
Mz y D
Iz
30 mm
z
(256.5 N m)( 0.030 m)
0.2344 10 6 m 4
32.83 MPa 17.40 MPa
50.23 MPa
25 mm
(704.8 N m)(0.025 m)
1.0125 10 6 m 4
D
50.2 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
596
A
PROBLEM 4.141
y
z
M ! 1.2 kN · m
10 mm
40 mm
The couple M acts in a vertical plane and is applied to a beam oriented
as shown. Determine the stress at point A.
10 mm
C
40 mm
70 mm
10 mm
Iy ! 1.894 # 106 mm4
Iz ! 0.614 # 106 mm4
Iyz ! $0.800 # 106 mm4
SOLUTION
Using Mohr’s circle, determine the principal axes and principal moments of inertia.
Y : (1.894, 0.800) 106 mm 4
Z : (0.614, 0.800) 106 mm 4
E : (1.254, 0) 106 mm 4
R
tan 2
EF
2
FZ
2
0.6402
0.8002 10
6
1.0245 106 mm 4
Iv
(1.254 1.0245) 106 mm 4
0.2295 106 mm 4
0.2295 10 6 m 4
Iu
(1.254 1.0245) 106 mm 4
2.2785 106 mm 4
2.2785 10 6 m 4
m
Mv
Mu
FZ 0.800 106
1.25
25.67
m
FE 0.640 106
M cos m (1.2 103 ) cos 25.67 1.0816 103 N m
M sin
(1.2 103 ) sin 25.67
m
0.5198 103 N m
uA
y A cos
m
z A sin
m
45 cos 25.67
45 sin 25.67
21.07 mm
vA
z A cos
m
y A sin
m
45 cos 25.67
45 sin 25.67
60.05 mm
A
M vuA
Iv
M u vA
Iu
(1.0816 103 )(21.07 10 3 )
0.2295 10 6
113.0 106 Pa
( 0.5198 103 )(60.05 10 3 )
2.2785 10 6
A
113.0 MPa
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597
PROBLEM 4.142
y
A
2.4 in.
The couple M acts in a vertical plane and is applied to a beam oriented as
shown. Determine the stress at point A.
2.4 in.
z
M ! 125 kip · in.
C
2.4 in.
2.4 in.
2.4 in. 2.4 in.
SOLUTION
Iy
2
1
(7.2)(2.4)3
3
66.355 in 4
Iz
2
1
(2.4)(7.2)3
12
(2.4)(7.2)(1.2) 2
I yz
2 (2.4)(7.2)(1.2)(1.2)
199.066 in 4
49.766 in 4
Using Mohr’s circle, determine the principal axes and principal moments of inertia.
Y : (66.355 in 4 , 49.766 in 4 )
Z : (199.066 in 4 ,
49.766 in 4 )
E : (132.710 in 4 , 0)
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598
PROBLEM 4.142 (Continued)
tan 2
m
2
m
R
DY 49.766
DE 66.355
36.87
18.435
m
DE
2
DY
2
82.944 in 4
Iu
132.710 82.944
49.766 in 4
Iv
132.710 82.944
215.654 in 4
Mu
125sin18.435
39.529 kip in.
Mv
125cos18.435
118.585 kip in.
uA
4.8 cos 18.435
2.4 sin 18.435
4.8 sin 18.435
A
M uA
I
A
Mu
Iu
2.4 cos 18.435
0.7589 in.
A
(118.585)(5.3126)
215.654
2.32 ksi
5.3126 in.
(39.529)(0.7589)
49.766
A
2.32 ksi
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599
y
PROBLEM 4.143
1.08 in.
0.75 in.
The couple M acts in a vertical plane and is applied to a beam oriented
as shown. Determine the stress at point A.
2.08 in.
z
C
M 5 60 kip · in.
6 in.
0.75 in.
A
4 in.
Iy 5 8.7 in4
Iz 5 24.5 in4
Iyz 5 18.3 in4
SOLUTION
Using Mohr’s circle, determine the principal axes and principal moments of inertia.
Y : (8.7, 8.3) in 4
Z : (24.5,
8.3) in 4
E : (16.6, 0) in 4
EF
7.9 in 4
FZ
8.3 in 4
R
7.92
8.32
11.46 in 4
tan 2
m
FZ 8.3
1.0506
EF 7.9
I v 16.6 11.46 28.06 in 4
23.2
Iu
16.6 11.46 5.14 in 4
Mu
M sin
m
(60) sin 23.2
23.64 kip in.
Mv
M cos
m
(60) cos 23.2
55.15 kip in.
uA
y A cos
m
z A sin
m
3.92 cos 23.2
1.08 sin 23.2
4.03 in.
vA
z A cos
m
y A sin
m
1.08cos 23.2
3.92 sin 23.2
0.552 in.
m
A
M vuA
Iv
M u vA
Iu
(55.15)( 4.03)
28.06
(23.64)(0.552)
5.14
A
10.46 ksi
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600
PROBLEM 4.144
D
B
H 14 kN
G
E
A
F
28 kN
The tube shown has a uniform wall thickness of 12 mm. For the
loading given, determine (a) the stress at points A and B, (b) the point
where the neutral axis intersects line ABD.
125 mm
75 mm
28 kN
SOLUTION
Add y- and z-axes as shown. Cross section is a 75 mm
cutout.
125-mm rectangle with a 51 mm
101-mm rectangular
1
1
(75)(125)3
(51)(101)3 7.8283 106 mm 4 7.8283 10 6 m 4
12
12
1
1
(125)(75)3
(101)(51)3 3.2781 103 mm 4 3.2781 10 6 m 4
Iy
12
12
A (75)(125) (51)(101) 4.224 103 mm 2 4.224 10 3 m 2
Iz
Resultant force and bending couples:
70 103 N
P 14 28 28 70 kN
(a)
A
Mz
(62.5 mm)(14 kN) (62.5 mm)(28kN) (62.5 mm)(28 kN)
2625 N m
My
(37.5 mm)(14 kN) (37.5 mm)(28 kN) (37.5 mm)(28 kN)
525 N m
P
A
M z yA
Iz
M y zA
70 103
4.224 10
Iy
3
(2625)( 0.0625)
7.8283 10 6
( 525)(0.0375)
3.2781 10 6
31.524 106 Pa
B
P
A
M z yB
Iz
A
M y zB
70 103
4.224 10
Iy
3
(2625)(0.0625)
7.8283 10 6
( 525)(0.0375)
3.2781 10 6
10.39 106 Pa
(b)
31.5 MPa
B
10.39 MPa
Let point H be the point where the neutral axis intersects AB.
zH
0
yH
0.0375 m,
P
A
yH
M z yH
Iz
Iz P
Mz A
?,
H
0
M y zH
Mz H
Iy
Iy
7.8283 10
2625
6
70 103
4.224 10
3
( 525)(0.0375)
3.2781 10 6
0.03151 m 31.51 mm
31.51 62.5 94.0 mm
Answer: 94.0 mm above point A.
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601
PROBLEM 4.145
A horizontal load P of magnitude 100 kN is applied to the beam
shown. Determine the largest distance a for which the maximum
tensile stress in the beam does not exceed 75 MPa.
y
20 mm
a
20 mm
O
z
x
P
20 mm
60 mm
20 mm
SOLUTION
Locate the centroid.
A, mm2
y , mm
2000
10
20 103
1200
3200
–10
12 103
Ay , mm3
Y
Ay
A
8 103
3200
2.5 mm
8 103
Move coordinate origin to the centroid.
Coordinates of load point:
XP
a,
Bending couples:
Mx
yP P
Ix
1
(100)(20)3
12
(2000)(7.5)2
yP
2.5 mm
My
1
(60)(20)3
12
aP
(1200)(12.5)2
0.40667 106 mm 4
0.40667 10
Iy
1
1
(20)(100)3
(20)(60)3
12
12
P M x y M yx
A
Ix
Iy
2.0267 106 mm 4
A
6
m4
2.0267 10 6 m 4
75 106 Pa, P
100 103 N
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602
PROBLEM 4.145 (Continued)
My
Iy P
x A
My
2.0267 10
50 10 3
Mxy
Ix
6
For point A,
100 103
3200 10 6
2.0267 10 6
{31.25
50 10 3
a
My
P
50 mm, y
( 2.5)(100 103 )( 2.5 10 3 )
0.40667 10 6
1.537
(1.7111 103 )
100 103
x
75} 106
2.5 mm
75 106
1.7111 103 N m
17.11 103 m
a
17.11 mm
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603
1 in.
PROBLEM 4.146
1 in.
4 in.
1 in.
5 in.
P
Knowing that P 90 kips, determine the largest distance a for
which the maximum compressive stress dose not exceed 18 ksi.
a
2.5 in.
SOLUTION
A
Ix
Iz
(5 in.)(6 in.)
2(2 in.)(4 in.)
14 in 2
1
1
(5 in.)(6 in.)3 2 (2 in.)(4 in.)3
12
12
1
1
2 (1 in.)(5 in.)3
(4 in.)(1 in.)3
12
12
21.17 in 4
Force-couple system at C:
P
P
For P
P
90 kips M x
90 kips:
Mx
68.67 in 4
Maximum compressive stress at B:
B
18 ksi
18
1.741
P
A
M x (3 in.)
Ix
90 kips
14 in 2
6.429
B
Mz
Pa
(90 kips)(2.5 in.)
225 kip in.
Mz
(90 kips) a
18 ksi
M z (2.5 in.)
Iz
(225 kip in.)(3 in.)
68.67 in 4
9.830
P(2.5 in.)
(90 kips) a (2.5 in.)
21.17 in 4
10.628a
10.628 a
a
0.1638 in.
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604
1 in.
PROBLEM 4.147
1 in.
4 in.
1 in.
P
5 in.
Knowing that a 1.25 in., determine the largest value of P that
can be applied without exceeding either of the following
allowable stresses:
a
2.5 in.
10 ksi
ten
18 ksi
comp
SOLUTION
A
Ix
Iz
(5 in.)(6 in.)
(2)(2 in.)(4 in.)
14 in 2
1
1
(5 in.)(6 in.)3 2 (2 in.)(4 in.)3 68.67 in 4
12
12
1
1
2 (1 in.)(5 in.)3
(4 in.)(1 in.)3 21.17 in 4
12
12
For a
Force-couple system at C:
P
My
P(2.5 in.)
P Mx
Pa
1.25 in.,
(1.25 in.)
Maximum compressive stress at B:
B
18 ksi
18
18
P
A
M x (3 in.)
Ix
P
14 in 2
0.0714P
0.3282P
P(2.5 in.)(3 in.)
68.67 in 4
0.1092 P
P
10 ksi
10
P
A
M x (3 in.)
Ix
0.0714P
0.1854 P
P(1.25 in.)(2.5 in.)
21.17 in 4
0.1476 P
54.8 kips
D
10 ksi
M z (2.5 in.)
Iz
0.1092 P
P
18 ksi
M z (2.5 in.)
Iz
Maximum tensile stress at D:
D
B
0.1476 P
53.9 kips
The smaller value of P is the largest allowable value.
P
53.9 kips
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605
y
R ! 125 mm
C
z
P ! 4 kN
PROBLEM 4.148
E
%
A rigid circular plate of 125-mm radius is attached to a solid
150 200-mm rectangular post, with the center of the plate directly above
30 ,
the center of the post. If a 4-kN force P is applied at E with
determine (a) the stress at point A, (b) the stress at point B, (c) the point
where the neutral axis intersects line ABD.
x
A
D
B
200 mm 150 mm
SOLUTION
4 103 N (compression)
P
Mx
Mz
Iz
xA
A
(a)
A
P
A
56.25 106 mm 4
100 106 mm 4
zA
zB
4 103
30 10 3
M z xA
Iz
56.25 10
100 10
6
6
B
P
A
M x zB
Ix
30 10
3
m2
( 433)( 100 10 3 )
100 10 6
( 250)(75 10 3 )
56.25 10 6
633 103 Pa
633 kPa
( 433)(100 10 3 )
100 10 6
B
(c)
m4
75 mm
( 250)(75 10 3 )
56.25 10 6
4 103
30 10 3
M z xB
Iz
433 N m
m4
A
(b)
250 N m
3
(4 10 )(125 10 ) cos 30
30 103 mm 2
(200)(150)
M xzA
Ix
3
PR cos 30
1
(200)(150)3
12
1
(150)(200)3
12
xB 100 mm
Ix
(4 103 )(125 10 3 )sin 30
PR sin 30
233 103 Pa
233 kPa
Let G be the point on AB where the neutral axis intersects.
G
0
P
A
G
xG
zG
75 mm
M x zG
Ix
Iz P
Mz A
46.2 10
M z xG
Iz
M x zG
Ix
3
m
xG
?
0
100 10
433
6
4 103
30 10 3
46.2 mm
( 250)(75 10 3 )
56.25 10 6
Point G lies 146.2 mm from point A.
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606
y
R ! 125 mm
C
z
P ! 4 kN
PROBLEM 4.149
E
%
In Prob. 4.148, determine (a) the value of θ for which the stress at D
reaches it largest value, (b) the corresponding values of the stress at A, B,
C, and D.
x
A
D
PROBLEM 4.148 A rigid circular plate of 125-mm radius is attached to a
solid 150 200-mm rectangular post, with the center of the plate directly
above the center of the post. If a 4-kN force P is applied at E with
30 , determine (a) the stress at point A, (b) the stress at point B,
(c) the point where the neutral axis intersects line ABD.
B
200 mm 150 mm
SOLUTION
(a)
4 103 N
P
PR sin
Mx
Iz
xD
For
M zx
Iz
d
d
to be a maximum,
d
d
D
P 0
RxD sin
IZ
tan
I z zD
I x xD
sin
A
P
A
0
Rz D cos
Ix
sin
cos
(b)
1
A
P
M xzA
Ix
M z xA
Iz
PR cos
Mx
30 103 mm 2
(200)(150)
A
M xz
Ix
500sin
500 N m
500 cos
1
(200)(150)3 56.25 106 mm 4 56.25 10 6 m 4
2
1
(150)(200)3 100 106 mm 4 100 10 6 m 4
2
100 mm
75 mm
zD
Ix
P
A
(4 103 )(125 10 3 )
PR
0.8,
Rz sin
Ix
Rx cos
Iz
with z
zD , x
30 10 3 m 2
xD
0
(100 10 6 )( 75 10 3 )
(56.25 10 6 )(100 10 3 )
cos
4 103
30 10 3
4
3
53.1
0.6
(500)(0.8)(75 10 3 )
56.25 10 6
(500)(0.6)( 100 10 3 )
100 10 6
( 0.13333
0.53333
0.300) 106 Pa
0.700 106 Pa
A
700 kPa
B
( 0.13333
0.53333
0.300) 106 Pa
0.100 106 Pa
B
100 kPa
C
( 0.13333
0
D
( 0.13333
0.53333
0) 106 Pa
133.3 kPa
C
0.300) 106 Pa
0.967 106 Pa
D
967 kPa
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607
y
PROBLEM 4.150
0.5 in.
1.43 in.
z
M0
C
5 in.
0.5 in.
1.43 in.
A beam having the cross section shown is subjected to a couple M 0 that
acts in a vertical plane. Determine the largest permissible value of the
moment M 0 of the couple if the maximum stress in the beam is
not to exceed 12 ksi. Given: I y I z 11.3 in 4 , A 4.75 in 2 ,
kmin 0.983 in. (Hint: By reason of symmetry, the principal axes form
2
an angle of 45 with the coordinate axes. Use the relations I min Akmin
and I min I max I y I z .)
5 in.
SOLUTION
Mu
M 0 sin 45
0.70711M 0
Mv
M 0 cos 45
0.70711M 0
I min
2
Akmin
I max
Iy
(4.75)(0.983)2
Iz
11.3
I min
11.3
uB
yB cos 45
z B sin 45
vB
z B cos 45
yB sin 45
B
M vu B
Iv
M u vB
Iu
0.70711M 0
M0
B
0.4124
4.59 in 4
3.57 cos 45
0.93 cos 45
0.70711M 0
( 1.866)
4.59
18.01 in 4
4.59
3.182
18.01
12
0.4124
uB
I min
0.93 sin 45
1.866 in.
( 3.57) sin 45
3.182 in.
vB
I max
0.4124M 0
M0
29.1 kip in.
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608
PROBLEM 4.151
y
0.5 in.
Solve Prob. 4.150, assuming that the couple M 0 acts in a horizontal
plane.
1.43 in.
z
M0
C
5 in.
0.5 in.
1.43 in.
5 in.
PROBLEM 4.150 A beam having the cross section shown is subjected
to a couple M 0 that acts in a vertical plane. Determine the largest
permissible value of the moment M 0 of the couple if the maximum
stress in the beam is not to exceed 12 ksi. Given: I y I z 11.3 in 4,
A 4.75 in 2, kmin 0.983 in. (Hint: By reason of symmetry, the
principal axes form an angle of 45 with the coordinate axes. Use the
2
relations I min Akmin
and I min I max I y I z .)
SOLUTION
Mu
Mv
M 0 cos 45
0.70711M 0
M 0 sin 45
I min
2
Akmin
I max
Iy
0.70711M 0
(4.75)(0.983)2
Iz
11.3
I min
4.59 in 4
11.3
4.59
18.01 in 4
uD
yD cos 45
z D sin 45
0.93 cos 45
( 3.57 sin 45 )
vD
z D cos 45
yD sin 45
( 3.57) cos 45
(0.93) sin 45
D
M vu D
Iv
M u vD
Iu
0.70711M 0
M0
D
0.4124
0.70711M 0
( 1.866)
4.59
3.182
18.01
12
0.4124
uD
I min
1.866 in.
3.182 in.
vD
I max
0.4124M 0
M0
29.1 kip in.
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609
PROBLEM 4.152
y
z
M0
40 mm
10 mm
C
10 mm
40 mm
70 mm
10 mm
The Z section shown is subjected to a couple M 0 acting in a vertical
plane. Determine the largest permissible value of the moment M 0 of the
couple if the maximum stress is not to exceed 80 MPa. Given:
I max 2.28 10 6 mm 4 ,
I min 0.23 10 6 mm 4 , principal axes
25.7
and 64.3
.
SOLUTION
Iv
I max
2.28 106 mm 4
2.28 10 6 m 4
Iu
I min
0.23 106 mm 4
0.23 10 6 m 4
Mv
M 0 cos 64.3
Mu
M 0 sin 64.3
64.3
tan
Iv
tan
Iu
2.28 10
0.23 10
20.597
6
6
tan 64.3
87.22
Points A and B are farthest from the neutral axis.
uB
yB cos 64.3
z B sin 64.3
( 45) cos 64.3
( 35) sin 64.3
yB sin 64.3
( 35) cos 64.3
( 45) sin 64.3
51.05 mm
vB
z B cos 64.3
25.37 mm
B
80 106
M vu B
Iv
M u vB
Iu
(M 0 cos 64.3 )( 51.05 10 3 )
2.28 10 6
(M 0 sin 64.3 )(25.37 10 3 )
0.23 10 6
109.1 103 M 0
M0
80 106
109.1 103
M0
733 N m
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610
PROBLEM 4.153
y
z
M0
40 mm
10 mm
C
10 mm
40 mm
70 mm
10 mm
Solve Prob. 4.152 assuming that the couple M 0 acts in a horizontal
plane.
PROBLEM 4.152 The Z section shown is subjected to a couple M 0
acting in a vertical plane. Determine the largest permissible value of the
moment M 0 of the couple if the maximum stress is not to exceed 80 MPa.
Given: I max 2.28 10 6 mm 4 , I min 0.23 10 6 mm 4 , principal axes
25.7
and 64.3
.
SOLUTION
Iv
I min
0.23 106 mm 4
0.23 106 m 4
Iu
I max
2.28 106 mm 4
2.28 106 m 4
Mv
M 0 cos 64.3
Mu
M 0 sin 64.3
64.3
tan
Iv
tan
Iu
0.23 10
2.28 10
0.20961
6
6
tan 64.3
11.84
Points D and E are farthest from the neutral axis.
uD
yD cos 25.7
z D sin 25.7
( 5) cos 25.7
yD sin 25.7
45cos 25.7
45 sin 25.7
24.02 mm
vD
z D cos 25.7
( 5) sin 25.7
38.38 mm
D
M vu D
Iv
M u vD
Iu
(M D cos 64.3 )( 24.02 10 3 )
0.23 10 6
(M 0 sin 64.3 )(38.38 10 3 )
2.28 10 6
80 106
M0
60.48 103 M 0
1.323 103 N m
M0
1.323 kN m
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611
PROBLEM 4.154
y
0.3 in.
M0
1.5 in.
C
z
0.3 in.
An extruded aluminum member having the cross section shown
is subjected to a couple acting in a vertical plane. Determine the
largest permissible value of the moment M 0 of the couple if the
maximum stress is not to exceed 12 ksi. Given:
I max 0.957 in 4 , I min 0.427 in 4 , principal axes 29.4
and
60.6
0.6 in. 1.5 in. 0.6 in.
SOLUTION
Iu
I max
0.957 in 4
Iv
I min
0.427 in 4
Mu
M 0 sin 29.4 , M v
M 0 cos 29.4
29.4
tan
Iv
tan
Iu
0.427
tan 29.4
0.957
0.2514
14.11
Point A is farthest from the neutral axis.
yA
0.75 in., z A
uA
y A cos 29.4
z A sin 29.4
1.0216 in .
vA
z A cos 29.4
y A sin 29.4
0.2852 in .
A
M vu A
Iv
M uVA
Iu
(M 0 cos 29.4 )( 1.0216)
0.427
0.75 in.
(M 0 sin 29.4 )( 0.2852)
0.957
1.9381 M 0
M0
A
1.9381
12
1.9381
M0
6.19 kip in.
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612
PROBLEM 4.155
y
20 mm
z
A beam having the cross section shown is subjected to a couple M0
acting in a vertical plane. Determine the largest permissible value of
the moment M0 of the couple if the maximum stress is not to exceed
100 MPa. Given: I y I z b 4 /36 and I yz b 4 /72.
M0
C
b = 60 mm
20 mm
b = 60 mm
SOLUTION
Iy
I yz
b 4 604
0.360 106 mm 4
36 36
b 4 604
0.180 106 mm 4
72 72
Iz
Principal axes are symmetry axes.
Using Mohr’s circle, determine the principal moments of inertia.
R
Iv
Iu
Mu
0.180 106 mm 4
I yz
Iy
Iz
R
2
0.540 106 mm 4
I y Iz
R
2
0.180 106 mm 4
M 0 sin 45
45
0.540 10 6 m 4
0.180 10 6 m 4
0.70711M 0 ,
Iv
tan
Iu
tan
Mv
0.540 10
0.180 10
M 0 cos 45
0.70711M 0
6
6
tan 45
3
71.56
Point A:
uA
A
M0
0
vA
M vuA
Iv
20 2 mm
M u vA
Iu
A
111.11 103
0
(0.70711M 0 )( 20 2 103 )
0.180 10
100 106
111.11 103
6
11.11 10 3 M 0
900 N m
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613
PROBLEM 4.155 (Continued)
Point B:
60
uB
2
mm,
M v uB
Iv
B
vB
M u vB
Iu
20
2
mm
(0.70711M 0 )
60
2
6
0.540 10
10
3
(0.70711M 0 )
20
2
0.180 10
10
3
6
111.11 103 M 0
M0
B
111.11 10
3
100 106
111.11 10
3
900 N m
M0
Choose the smaller value.
900 N m
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614
b
PROBLEM 4.156
B
A
Show that, if a solid rectangular beam is bent by a couple applied in a plane containing
one diagonal of a rectangular cross section, the neutral axis will lie along the other
diagonal.
h
M
C
D
E
SOLUTION
tan
Mz
Iz
tan
b
h
M cos , M z
1 3
bh
12
Iz
tan
Iy
M sin
Iy
1 3
hb
12
1 3
bh b
12
1 3 h
hb
12
h
b
The neutral axis passes through corner A of the diagonal AD.
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615
PROBLEM 4.157
y
(a) Show that the stress at corner A of the prismatic
member shown in part a of the figure will be zero if
the vertical force P is applied at a point located on the
line
D
A
A
D
P
B
z
x
z
C
h
6
x
x
b /6
h
B
b
(a)
C
1
(b) Further show that, if no tensile stress is to occur in
the member, the force P must be applied at a point
located within the area bounded by the line found in
part a and three similar lines corresponding to the
condition of zero stress at B, C, and D, respectively.
This area, shown in part b of the figure, is known as
the kern of the cross section.
b
6
(b)
z
h /6
SOLUTION
1 3
hb
Ix
12
h
xA
2
Iz
zA
1 3
bh
12
b
2
A
bh
Let P be the load point.
Mz
A
x
b/6
z
h/6
0,
1
At point E:
z
0
xE
b /6
At point F:
x
0
zF
h/6
(a)
For
(b)
A
0
PxP
Mx
Pz P
P
A
M z xA
Iz
P
bh
( PxP )
1
12
M xzA
Ix
hb
b
2
3
( Pz P
1 bh
12
P
1
bh
xP
b /6
zP
h/6
x
b/6
z
h/6
1
3
h)
2
If the line of action ( xP , z P ) lies within the portion marked TA , a tensile stress will occur at corner A.
By considering
B
0,
C
0, and
D
0, the other portions producing tensile stresses are identified.
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616
PROBLEM 4.158
y
A beam of unsymmetric cross section is subjected to a couple M 0 acting in the
horizontal plane xz. Show that the stress at point A, of coordinates y and z, is
z
A
y
C
z
A
zI z
I yIz
yI yz
2
I yz
My
x
where Iy, Iz, and Iyz denote the moments and product of inertia of the cross section
with respect to the coordinate axes, and My the moment of the couple.
SOLUTION
The stress A varies linearly with the coordinates y and z. Since the axial force is zero, the y- and z-axes are
centroidal axes:
A
C1 y
C2 z
where C1 and C2 are constants.
Mz
y
I zC1
C1
My
I yz
Iz
z
I yzC2
AdA
I yz
C2
C1
A
C2 yz dA
0
C2
I yzC1
IzM y
C1 y 2dA
AdA
C1 yz dA
C2 z 2dA
I y C2
I yz
Iz
C2
I y C2
2
I yz
C2
I yIz
IzM y
I yIz
2
I yz
I yz M y
I yIz
Izz
I yIz
2
I yz
I yz y
2
I yz
My
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617
PROBLEM 4.159
y
A beam of unsymmetric cross section is subjected to a couple M 0 acting in the
vertical plane xy. Show that the stress at point A, of coordinates y and z, is
z
A
y
C
z
A
yI y
zI yz
I yIz
2
I yz
Mz
x
where I y , I z , and I yz denote the moments and product of inertia of the cross
section with respect to the coordinate axes, and M z the moment of the couple.
SOLUTION
The stress A varies linearly with the coordinates y and z. Since the axial force is zero, the y- and z-axes are
centroidal axes:
A
C1 y
C2 z
where C1 and C2 are constants.
My
z
AdA
I yzC1
C2
Mz
I yz
Iy
I y C2
C1
C2
A
C2 z 2dA
0
C1
y
C1 y 2dA
Adz
I z C1
I yM z
C1 yz dA
I yz
I yz
Iy
C2 yz dA
C1
2
I yz
C1
I yIz
I yM z
I yIz
2
I yz
I yz M z
I yIz
2
I yz
Iyy
I yz 2
I yIz
2
I yz
Mz
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618
PROBLEM 4.160
y
(a) Show that, if a vertical force P is applied at point A of the section
shown, the equation of the neutral axis BD is
D
B
P
C
A
xA
z
xA
x
rz2
x
zA
zA
z
rx2
1
where rz and rx denote the radius of gyration of the cross section with
respect to the z axis and the x axis, respectively. (b) Further show that, if
a vertical force Q is applied at any point located on line BD, the stress at
point A will be zero.
SOLUTION
Definitions:
Ix
, rz2
A
rx2
(a)
Mx
Pz A
E
P
A
Mz
Px A
M z xE
Iz
M x zE
Ix
P
1
A
xA
xE
rz2
P
A
zA
zE
rx2
Px A xE
Arz2
Iz
A
Pz A z E
Arx2
0
if E lies on neutral axis.
1
(b)
Mx
Pz E
A
P
A
Mz
M z xA
Iz
xA
x
rz2
zA
z
rx2
0,
xA
x
rz2
P
A
PxE x A
Arz2
Pz E z A
Arx2
zA
z
rx2
1
PxE
M x zA
Iy
0 by equation from part (a).
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619
24 mm
B
PROBLEM 4.161
B
For the curved bar shown, determine the stress at point A when
(a) h 50 mm, (b) h 60 mm.
h
A
600 N · m
A
50 mm
600 N · m
C
SOLUTION
(a)
h
50 mm,
A
(24)(50)
h
r
ln 2
r1
R
r
e
yA
A
r1
50 mm,
r2
100 mm
1.200 103 mm
50
ln 100
50
72.13475 mm
1
(r1 r2 ) 75 mm
2
r R 2.8652 mm
72.13475
50
22.13475 mm
rA
50 mm
3
My A
AerA
(600)(22.13475 10 )
(1.200 10 3 )(2.8652 10 3 )(50 10 3 )
77.3 106 Pa
77.3 MPa
(b)
h
R
r
yA
A
60 mm,
h
r
ln r2
1
r1
60
ln 110
50
50 mm,
r2
110 mm,
A
(24)(60)
1.440 103 mm 2
76.09796 mm
1
(r1 r2 ) 80 mm
e r R 3.90204 mm
2
76.09796 50 26.09796 mm
rA 50 mm
M yA
AerA
(600)(26.09796 10 3 )
(1.440 10 3 )(3.90204 10 3 )(50 10 3 )
55.7 106 Pa
55.7 MPa
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620
24 mm
B
B
A
A
PROBLEM 4.162
For the curved bar shown, determine the stress at points A and B
when h 55 mm.
h
600 N · m
C
50 mm
600 N · m
SOLUTION
h
A
R
r
e
yA
A
55 mm,
(24)(55)
r1
50 mm,
3
1.320 10 mm
r2
105 mm
2
50
h
74.13025 mm
105
r2
ln
ln
50
r1
1
(r1 r2 ) 77.5 mm
2
r R 3.36975 mm
74.13025
M yA
AerA
50
24.13025 mm
rA
50 mm
(600)(24.13025 10 3 )
(1.320 10 3 )(3.36975 10 3 )(50 10 3 )
65.1 106 Pa
65.1 MPa
yB
B
74.13025
MyB
AerB
105
30.86975 mm
rB
105 mm
3
(600)( 30.8697 10 )
(1.320 10 3 )(3.36975 10 3 )(105 10 3 )
39.7 106 Pa
39.7 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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621
4 kip · in.
PROBLEM 4.163
C
4 kip · in.
3 in.
For the machine component and loading shown, determine the
stress at point A when (a) h 2 in., (b) h 2.6 in.
h
A
0.75 in.
B
SOLUTION
M
4 kip in.
A
Rectangular cross section:
(a)
r
1
(r1
2
h
2 in.
r1
3
R
2
ln 13
2
(b)
M (r R )
Aer
h
2.6 in.
r1
3
R
2.6
3
ln 0.4
2.6
r1
e
2
r
r1
1)
1.8205
2 in.
0.1795 in.
r
e
12.19 ksi
A
12.19 ksi
A
11.15 ksi
1.95 in 2
0.4 in.
M (r R )
Aer
1
(3
2
1 in.
(0.75)(2.6)
1.2904 in.
h
R
( 4)(1 1.8205)
(1.5)(0.1795)(1)
A
At point A:
A
r
1.8205 in.
r2
1.5 in 2
(0.75)(2)
r
3 in. r1
r
1 in.
At point A:
A
h
, e
r
ln 2
r1
r2 ), R
A
bh r2
1.7
1
(3
2
0.4)
1.2904
1.7 in.
0.4906 in.
0.4 in.
( 4)(0.4 1.2904)
(1.95)(0.4096)(0.4)
11.15 ksi
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622
4 kip · in.
4 kip · in.
PROBLEM 4.164
C
3 in.
For the machine component and loading shown, determine the
stress at points A and B when h 2.5 in.
h
A
0.75 in.
B
SOLUTION
M
4 kip in.
Rectangular cross section:
h
2.5 in. b
r2
3 in. r1
r
R
At point A:
At point B:
1
(r1
2
h
ln
r2
r2 )
2.5
ln 3.0
0.5
r2
r1
1.75
e
r
R
r
r1
0.5 in.
A
M (r R)
Aer
r
r2
B
M (r R )
Aer
1.875 in.2
0.75 in. A
h
0.5 in.
1
(0.5
2
3.0)
1.75 in.
1.3953 in.
1.3953
0.3547 in.
( 4 kip in.)(0.5 in. 1.3953 in.)
(0.75 in.)(2.5 in.)(0.3547 in.)(0.5 in.)
A
10.77 ksi
( 4 kip in.)(3 in. 1.3953 in.)
(0.75 in. 2.5 in.)(0.3547 in.)(3 in.)
B
3.22 ksi
3 in.
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623
PROBLEM 4.165
40 mm
r1
The curved bar shown has a cross section of 40 60 mm and an inner
radius r1 15 mm. For the loading shown, determine the largest tensile
and compressive stresses.
60 mm
120 N · m
SOLUTION
h
40 mm, r1
A
(60)(40)
R
At r
h
r
ln 2
r1
r
1
(r1
2
e
r
2400 mm 2
40
55
ln
40
r2 )
R
15 mm,
15 mm, r2
55 mm
2400 10 6 m 2
30.786 mm
35 mm
My
Aer
4.214 mm
y
30.786
15
15.756 mm
(120)(15.786 10 3 )
(2400 10 6 )(4.214 10 3 )(15 10 3 )
12.49 10
6
Pa
12.49 MPa
At r
55 mm,
y
30.786
55
24.214 mm
(120)( 24.214 10 3 )
(2400 10 6 )(4.214 10 3 )(55 10 3 )
5.22 106 Pa
5.22 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
624
PROBLEM 4.166
40 mm
r1
For the curved bar and loading shown, determine the percent
error introduced in the computation of the maximum stress by
assuming that the bar is straight. Consider the case when
(a) r1 20 mm, (b) r1 200 mm, (c) r1 2 m.
60 mm
120 N · m
SOLUTION
h
40 mm, A
I
1 3
bh
12
(60)(40)
1
(60)(40)3
12
2400 mm 2
2400 10
0.32 106 mm 4
6
m2 , M
120 N m
0.32 10 6 mm 4 ,
c
1
h
2
20 mm
Assuming that the bar is straight,
s
(a)
r1
R
(120)(20 10 8 )
(0.32 10 6 )
Mc
I
20 mm r2
h
r
ln 2
r1
1
(r1
2
a
M (r1 R)
Aer
r2 )
% error
7.5 MPa
60 mm
40
60
ln
20
r
7.5 106 Pa
36.4096 mm
e
40 mm
r1
R
r
R
16.4096 mm
3.5904 mm
(120)( 16.4096 10 3 )
(2400 10 6 )(3.5904 10 3 )(20 10 3 )
11.426 ( 7.5)
11.426
100%
11.426 106 Pa
11.426 MPa
34.4%
For parts (b) and (c), we get the values in the table below:
(a)
(b)
(c)
r1, mm
r2 , mm
20
200
2000
60
240
2040
R, mm
36.4096
219.3926
2019.9340
r , mm
40
220
2020
e, mm
3.5904
0.6074
0.0660
, MPa
11.426
7.982
7.546
% error
34.4 %
6.0 %
0.6 %
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
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625
0.3 in.
B
0.4 in.
P9
PROBLEM 4.167
B
P
0.4 in.
Steel links having the cross section shown are available with
different central angles . Knowing that the allowable stress is
12 ksi, determine the largest force P that can be applied to a
link for which
90 .
0.8 in.
A
A
0.8 in.
b
1.2 in.
C
C
SOLUTION
Reduce section force to a force-couple system at G, the centroid of the cross section AB.
a
The bending couple is M
r 1
cos
2
Pa.
For the rectangular section, the neutral axis for bending couple only lies at
h
R
Also, e
r
.
r2
r1
ln
R
At point A, the tensile stress is
where
A
K
1
r
1.2 in., r1
A
(0.3)(0.8)
e
1.2
a
1.2(1
K
1
P
(0.24)(12)
4.3940
My A
Aer1
ay A
er1
P
A
Pay A
Aer1
yA
and
R
P
1
A
ay A
er1
K
P
A
r1
A A
K
P
Data:
P
A
0.8 in., r2
1.6 in., h
0.24 in 2
1.154156
cos 45 )
R
0.045844 in.,
yA
0.8 in., b 0.3 in.
0.8
1.154156 in.
ln 1.6
0.8
1.154156
0.8
0.35416 in.
0.35147 in.
(0.35147)(0.35416)
(0.045844)(0.8)
4.3940
P
0.65544 kips
655 lb
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
626
PROBLEM 4.168
0.3 in.
B
B
0.4 in.
P9
P
0.4 in.
A
Solve Prob. 4.167, assuming that
PROBLEM 4.167 Steel links having the cross section shown are
available with different central angles . Knowing that the
allowable stress is 12 ksi, determine the largest force P that can be
applied to a link for which
90 .
A
0.8 in.
b
1.2 in.
C
60 .
0.8 in.
C
SOLUTION
Reduce section force to a force-couple system at G, the centroid of the cross section AB.
a
The bending couple is M
r 1
cos
2
Pa.
For the rectangular section, the neutral axis for bending couple only lies at
h
R
Also, e
r
r2
r1
.
R
At point A, the tensile stress is
r
1.2 in., r1
A
(0.3)(0.8)
e
1.2
a
(1.2)(1
K
1
P
(0.24)(12)
2.5525
P
A
A
where
Data:
ln
My A
Aer1
ay A
er1
K
1
P
A A
K
0.8 in., r2
P
A
and
1.6 in., h
0.24 in 2
1.154156
cos 30 )
0.045844 in.
Pay A
Aer1
P
1
A
yA
R
ay A
er1
K
P
A
r1
0.8 in., b 0.3 in.
0.8
1.154156 in.
R
ln 1.6
0.8
yA
1.154156
0.8
0.35416 in.
0.160770 in.
(0.160770)(0.35416)
(0.045844)(0.8)
2.5525
P
1.128 kips
1128 lb
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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627
5 kN
PROBLEM 4.169
a
The curved bar shown has a cross section of 30 30 mm. Knowing
that the allowable compressive stress is 175 MPa, determine the largest
allowable distance a.
30 mm
B
A
20 mm
20 mm
C
30 mm
5 kN
SOLUTION
Reduce the internal forces transmitted across section AB to a force-couple system at the centroid of the
section. The bending couple is
M
P(a
r)
Also, e
r
h
.
r2
ln
r1
R
For the rectangular section, the neutral axis for bending couple only lies at
R
The maximum compressive stress occurs at point A . It is given by
P
A
A
K
Thus, K
Data:
h
30 mm, r1
e
35
a
r,
50 mm, r
2.2593 mm, b
6
175 10 Pa, P
6
a
r
r
(30.5)(2.2593)(20)
12.7407
a
108.17 mm
P
A
35 mm, R
5 kN
P (a
P
with y A R
A
(a r )( R r1)
er1
30 mm, R
r1
30
ln 50
20
r ) yA
Aer1
r1
(1)
32.7407 mm
12.7407 mm, a
?
3
5 10 N
6
(900 10 )( 175 10 )
5 103
A A
P
K
a
32.7407
175 MPa
A
Solving (1) for
20 mm, r2
1
My A
Aer1
31.5
( K 1)er1
R r1
108.17 mm
35 mm
a
73.2 mm
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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628
PROBLEM 4.170
2500 N
For the split ring shown, determine the stress at (a) point A,
(b) point B.
90 mm
40 mm
B
A
14 mm
SOLUTION
r1
1
40
2
20 mm,
r2
A (14)(25) 350 mm 2
1
(90) 45 mm
2
25
h
R
r2
45
ln r ln 20
h
r2
r1
25 mm
30.8288 mm
1
r
1
(r1
2
r2 ) 32.5 mm
e
r
R 1.6712 mm
Reduce the internal forces transmitted across section AB to a force-couple system at the centroid of the cross
section. The bending couple is
M
(a)
Pa
rA
Point A:
A
P
A
(2500)(32.5 10 3 ) 81.25 N m
Pr
20 mm
My A
AeR
yA
2500
350 10
30.8288 20 10.8288 mm
(81.25)(10.8288 10 3 )
(350 10 6 )(1.6712 10 3 )(20 10 3 )
6
82.4 106 Pa
(b)
Point B:
B
P
A
rB
45 mm
MyB
AerB
82.4 MPa
A
yB
2500
350 10
30.8288 45
6
14.1712 mm
(81.25)( 14.1712 10 3 )
(350 10 6 )(1.6712 10 3 )(45 10 3 )
36.6 106 Pa
B
36.6 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
629
2 in.
PROBLEM 4.171
B
0.5 in.
0.5 in.
2 in.
0.5 in.
A
3 in.
M
M'
3 in.
Three plates are welded together to form the curved
beam shown. For M 8 kip in., determine the stress
at (a) point A, (b) point B, (c) the centroid of the cross
section.
C
SOLUTION
A
dA
R
1
r
Part
b
3.5
5.5
6
(a)
yA
R
yB
R
yC
C
R
3.25
4.875
1.0
0.225993
4.5
4.5
1.0
0.174023
5.75
5.75
3.5
0.862468
0.5
1.5
0.5
2
2
0.5
b ln
3.5
4.05812 in.,
0.862468
r R 0.26331 in.
3
r
M
15.125
15.125
4.32143 in.
3.5
8 kip in.
1.05812 in.
( 8)(1.05812)
(3.5)(0.26331)(3)
4.05812
MyB
Aer2
B
(c)
r2
0.462452
3
4.05812
My A
Aer1
A
(b)
r1
Ar
A
e
ri 1
ri
r
h
R
A
r
bi ln i 1
ri
Ari
A
r
r
3
bi hi
r
bi ln i 1
ri
r
MyC
Aer
6
A
3.06 ksi
1.94188 in.
( 8)( 1.94188)
(3.5)(0.26331)(6)
B
2.81 ksi
C
0.529 ksi
e
Me
Aer
M
Ar
8
(3.5)(4.32143)
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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630
2 in.
PROBLEM 4.172
B
0.5 in.
0.5 in.
2 in.
0.5 in.
A
3 in.
M
M'
3 in.
Three plates are welded together to form the curved beam
shown. For the given loading, determine the distance e
between the neutral axis and the centroid of the cross
section.
C
SOLUTION
A
dA
R
b
Ar
0.462452
3.25
4.875
1.0
0.225993
4.5
4.5
1.0
0.174023
5.75
5.75
3.5
0.862468
A
3
0.5
1.5
5.5
0.5
2
6
2
0.5
3.5
R
e
ri 1
ri
r
h
r
3
A
r
bi ln i 1
ri
Ari
A
r
Part
bi hi
r
bi ln i 1
ri
1
r
b ln
3.5
4.05812 in., r
0.862468
r R 0.26331 in.
15.125
3.5
15.125
4.32143 in.
e
0.263 in.
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
631
PROBLEM 4.173
C
M
A
Mʹ
150 mm
A
Knowing that the maximum allowable stress is 45 MPa, determine
the magnitude of the largest moment M that can be applied to the
components shown.
45 mm
135 mm
B
B
36 mm
SOLUTION
A
dA
R
1
r
bi hi
r
bi 2 ln i 1
ri
Ai
r
bi ln i 1
ri
Ai ri
Ai
r
r, mm
150
Part bi , mm h, mm
195
330
R
e
yA
r1
ri 1
, mm
ri
r , mm
Ar , mm3
108
45
4860
28.3353
172.5
838.35 103
36
135
4860
18.9394
262.5
1275.75 103
9720
47.2747
r
2114.1 103
9720
9720
205.606 mm
47.2747
r R 11.894 mm
R
bi ln
A, mm 2
205.606
150
55.606 mm
A Aer1
M
yA
(45 106 )(9720 10 6 )(11.894 10 3 )(150 10 3 )
(55.606 10 3 )
B
M
217.5 mm
My A
Aer1
A
yB
2114.1 103
R
r2
205.606
330
14.03 kN m
124.394 mm
MyB
Aer2
B Aer2
yB
(45 106 )(9720 10 6 )(11.894 10 3 )(330 10 3 )
(124.394 10 3 )
13.80 kN m
M
13.80 kN m
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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632
C
M
A
PROBLEM 4.174
Mʹ
150 mm
A
B
B
Knowing that the maximum allowable stress is 45 MPa, determine
the magnitude of the largest moment M that can be applied to the
components shown.
135 mm
45 mm
36 mm
SOLUTION
A
dA
R
1
r
bi hi
r
bi ln i 1
ri
Ai
r
bi ln i 1
ri
Ai ri
Ai
r
bi , mm h, mm
r, mm
150
285
bi ln
A, mm 2
ri 1
, mm
ri2
135
4860
23.1067
217.5
1.05705 106
108
45
4860
15.8332
307.5
1.49445 106
9720
38.9399
R
9720
38.9399
e
r
R
12.885 mm,
yA
R
r1
249.615
249.615 mm,
M
150
2.5515 106
9720
r
262.5 mm
20 103 N m
99.615 mm
A Aer1
M
yA
(45 106 )(9720 10 6 )(12.885 10 3 )(150 10 3 )
(99.615 10 3 )
M
2.5515 106
My A
Aer1
A
B
Ar , mm3
36
330
yB
r , mm
R
r2
249.615
330
8.49 kN m
80.385 mm
MyB
Aer2
B Aer2
yB
(45 106 )(9720 10 6 )(12.885 10 3 )(330 10 3 )
(80.385 10 3 )
23.1 kN m
M
8.49 kN m
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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633
120 lb
PROBLEM 4.175
120 lb
The split ring shown has an inner radius r1 0.8 in. and a circular cross
section of diameter d 0.6 in. Knowing that each of the 120-lb forces is
applied at the centroid of the cross section, determine the stress (a) at point A,
(b) at point B.
r1
d
A
B
SOLUTION
rA
r1
0.8 in.
rB
rA
d
r
1
(rA
2
A
c2
e
1
r
2
r R
Q
120 lb
R
M0
(a)
0.8
rB )
0.6
1.4 in.
1.1 in.
(0.3)2
c
1
d
2
0.28274 in 2
0.3 in.
for solid circular section
1
1.1
(1.1)2 (0.3)2
2
1.1 1.0791503 0.020850 in.
r2
c2
Fx
0:
2rQ
r
rA
0.8 in.
A
P
A
M (rA R)
AerA
M
0: P
0
120
0.28274
Q
M
1.079150 in.
0
P
120 lb
2rQ
(2)(1.1)(120)
( 264)(0.8 1.079150)
(0.28274)(0.020850)(0.8)
264 lb in.
16.05 103 psi
A
(b)
r
rB
1.4 in.
B
P
A
M (rB R)
AerB
120
0.28274
16.05 ksi
( 264)(1.4 1.079150)
(0.28274)(0.020850)(1.4)
9.84 103 psi
B
9.84 ksi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
634
120 lb
PROBLEM 4.176
120 lb
Solve Prob. 4.175, assuming that the ring has an inner radius r1 = 0.6 in. and
a cross-sectional diameter d 0.8 in.
r1
PROBLEM 4.175 The split ring shown has an inner radius r1 0.8 in. and a
circular cross section of diameter d 0.6 in. Knowing that each of the 120-lb
forces is applied at the centroid of the cross section, determine the stress
(a) at point A, (b) at point B.
d
A
B
SOLUTION
rA
r1
0.6 in.
rB
rA
d
r
1
(rA
2
A
c2
e
1
r
2
r R
Q
120 lb
R
0:
M0
(a)
0.6
rB )
0.8
1.0 in.
(0.4)2
r2
1.0
1.4 in.
c
0.50265 in 2
0.4 in.
for solid circular section.
1
1.0
(1.0) 2 (0.4)2
2
0.958258 0.041742 in.
c2
Fx
2rQ
1
d
2
M
r
rA
0.6 in.
A
P
A
M (rA R)
AerA
0
0
P
M
120
0.50265
Q
2rQ
0
0.958258 in.
P
120 lb
(2)(1.0)(120)
240 lb in.
( 240)(0.6 0.958258)
(0.50265)(0.041742)(0.6)
7.069 103 psi
(b)
r
rB
1.4 in.
B
P
A
M (rB R)
AerB
A
120
0.50265
7.07 ksi
( 240)(1.4 0.958258)
(0.50265)(0.041742)(1.4)
3.37 103 psi
B
3.37 ksi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
635
220 N
B
PROBLEM 4.177
The bar shown has a circular cross section of 14-mm diameter. Knowing
that a 32 mm, determine the stress at (a) point A, (b) point B.
A C
220 N
a
16 mm
12 mm
SOLUTION
c
R
e
(a)
1
d 8 mm
r 12 8 20 mm
2
1
1
r
r 2 c2
20
202 82
2
2
r R 20 19.1652 0.83485 mm
A
r2
(8) 2
P
220 N
M
P (a
201.06 mm 2
r)
220(0.032
yA
R
r1
19.1652
12
yb
R
r2
19.1652
28
A
P
A
My A
Aer1
220
201.06 10
19.1652 mm
6
0.020)
11.44 N m
7.1652 mm
8.8348 mm
( 11.44)(7.1652 10 3 )
(201.06 10 6 )(0.83485 10 3 )(12 10 3 )
A
(b)
B
P
A
41.8 MPa
MyB
Aer2
220
201.06 10
6
( 11.44)(8.8348 10 3 )
(201.06 10 6 )(0.83485 10 3 )(28 10 3 )
B
20.4 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
636
220 N
B
PROBLEM 4.178
The bar shown has a circular cross section of 14-mm diameter. Knowing
that the allowable stress is 38 MPa, determine the largest permissible
distance a from the line of action of the 220-N forces to the plane
containing the center of curvature of the bar.
A C
220 N
a
16 mm
12 mm
SOLUTION
c
R
e
1
d 8 mm
r 12 8 20 mm
2
1
1
r
r 2 c2
20
202 82
2
2
r R 20 19.1652 0.83485 mm
A
r2
(8) 2
P
220 N
M
P (a
R
r1
A
P
A
My A
Aer1
K
a
r
201.06 mm 2
r)
yA
KP
A
19.1652 mm
19.1652
P
A
where
12
P(a
K
7.1652 mm
r ) yA
Aer1
P
1
A
(a
r ) yA
er1
1
(38 106 )(201.06 10 6 )
P
220
( K 1)er1
yA
AA
(34.729
(a
r ) yA
er1
34.729
1)(0.83485 10 3 )(12 10 3 )
(7.1652 10 3 )
0.047158 m
a
0.047158
0.020
0.027158
a
27.2 mm
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
637
C
M
PROBLEM 4.179
16 mm
The curved bar shown has a circular cross section of 32-mm
diameter. Determine the largest couple M that can be applied to
the bar about a horizontal axis if the maximum stress is not to
exceed 60 MPa.
12 mm
SOLUTION
c
R
e
max
16 mm
r
1
r
2
1
28
2
r2
r
28
16
28 mm
c2
282
R
12
162
25.4891 mm
25.4891
2.5109 mm
occurs at A, which lies at the inner radius.
It is given by
Also,
A
Data:
yA
M
My A
from which
Aer1
max
c2
R
(16) 2
r1
M
Aer1
yA
max
.
804.25 mm 2
25.4891 12
13.4891 mm
(804.25 10 6 )(2.5109 10 3 )(12 10 3 )(60 106 )
13.4891 10 3
M
107.8 N m
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
638
P
90 mm
PROBLEM 4.180
Knowing that P
B
10 kN, determine the stress at (a) point A, (b) point B.
80 mm
A
100 mm
SOLUTION
Locate the centroid D of the cross section.
r
90 mm
3
100 mm
130 mm
Force-couple system at D.
P
M
10 kN
Pr
1300 N m
(10 kN)(130 mm)
Triangular cross section.
A
1
bh
2
1
(90 mm)(80 mm)
2
3600 mm 2
R
(a)
A
Point A:
P
A
M (rA R)
AerA
2.778 MPa
(b)
B
Point B:
P
A
rA
1
R
126.752 mm
e
r
R
1
(90)
2
190 190
ln
90 100
130 mm
1
45 mm
0.355025
126.752 mm
3.248 mm
0.100 m
10 kN
3600 10 6 m 2
(1300 N m)(0.100 m 0.126752 m)
(3600 10 6 m 2 )(3248 10 3 m)(0.100 m)
29.743 MPa
rB
M (rB R)
AerB
2.778 MPa
100 mm
1
h
2
r2 r2
ln
h r1
3600 10 6 m 2
190 mm
A
32.5 MPa
B
34.2 MPa
0.190 m
10 kN
3600 10 6 m 2
(1300 N m)(0.190 m 0.126752 m)
(3600 10 6 m 2 )(3.248 10 3 m)(0.190 m)
37.01 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
639
M
2.5 in.
A
PROBLEM 4.181
B
M
Knowing that M
(b) point B.
C
3 in.
2 in.
2 in.
5 kip in., determine the stress at (a) point A,
3 in.
SOLUTION
r
1
bh
2
2 1
b1
2.5 in., r1
A
Use formula for trapezoid with
b2
1
(2.5)(3) 3.75 in 2
2
3.00000 in.
1 h 2 (b b )
1
2
2
r2
b2r1) ln
h(b1
r1
(b1r2
(a)
yA
R
yB
B
R
5 in.
b2 )
(0.5)(3)2 (2.5
[(2.5)(5) (0)(2)] ln 52
0)
(3)(2.5
r
M
R
0.15452 in.
0)
2.84548 in.
5 kip in.
0.84548 in.
My A
Aer1
A
(b)
r1
0, r2
0.
R
e
2 in., b2
r2
MyB
Aer2
(5)(0.84548)
(3.75)(0.15452)(2)
3.65 ksi
A
2.15452 in.
(5)( 2.15452)
(3.75)(0.15452)(5)
B
3.72 ksi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
640
M
PROBLEM 4.182
B
2.5 in.
A
3 in.
Knowing that M
(b) point B.
M
5 kip in., determine the stress at (a) point A,
C
2 in.
2 in.
3 in.
SOLUTION
r
1 (2.5)(3) 3.75 in 2
2
2 2 4.00000 in.
b1
0, r1
b1
0.
A
Use formula for trapezoid with
2 in.,
b2
1 h 2 (b b )
1
2
2
r2
b2r1) ln
h (b1
r1
R
(b1r2
(0.5)(3) 2 (0
[(0)(5)
e
(a)
yA
R
yB
B
R
R
(2.5)(2)] ln 5
2
0.14534 in.
r2
5 in.
b2 )
2.5)
(3)(0
M
2.5)
3.85466 in.
5 kip in.
1.85466 in.
My A
Aer1
A
(b)
r1
r
2.5 in.,
r2
MyB
Aer2
(5)(1.85466)
(3.75)(0.14534)(2)
A
8.51 ksi
1.14534 in.
(5)( 1.14534)
(3.75)(0.14534)(5)
B
2.10 ksi
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
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641
PROBLEM 4.183
80 kip · in.
b
B
A
B
A
Knowing that the machine component shown has a trapezoidal cross
section with a 3.5 in. and b 2.5 in., determine the stress at
(a) point A, (b) point B.
C
a
6 in. 4 in.
SOLUTION
Locate centroid.
A, in 2
r , in.
Ar , in 3
10.5
6
63
7.5
8
60
18
r
R
123
123
18
6.8333 in.
1
2
(b1r2
h 2 (b1
b2 r1 ) ln
b2 )
r2
r1
h(b1 b2 )
(0.5)(6)2 (3.5 2.5)
[(3.5)(10) (2.5)(4)]ln 104 (6)(3.5 2.5)
e
(a)
yA
A
(b)
yB
B
r
R
0.4452 in.
M
6.3878 in.
80 kip in.
R r1 6.3878 4 2.3878 in.
MyA
(80)(2.3878)
(18)(0.4452)(4)
Aer1
5.96 ksi
A
R r2 6.3878 10
3.6122 in.
MyB
(80)( 3.6122)
(18)(0.4452)(10)
Aer2
B
3.61 ksi
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642
PROBLEM 4.184
80 kip · in.
b
B
A
B
A
Knowing that the machine component shown has a trapezoidal cross
section with a 2.5 in. and b 3.5 in., determine the stress at
(a) point A, (b) point B.
C
a
6 in. 4 in.
SOLUTION
Locate centroid.
A, in 2
r , in.
Ar , in 3
7.5
6
45
10.5
8
84
18
r
R
e
(a)
yA
A
(b)
yB
B
R r1
129
18
7.1667 in.
1
2
(b1r2
h 2 (b1 b2 )
b2 r1 ) ln
r2
r1
h(b1 b2 )
(0.5)(6) 2 (2.5 3.5)
[(2.5)(10) (3.5)(4)]ln 104 (6)(2.5 3.5)
6.7168 in.
r
80 kip in.
R
M
0.4499 in.
2.7168 in.
My A
Aer1
R r2
129
(80)(2.7168)
(18)(0.4499)(4)
A
6.71 ksi
3.2832 in.
MyB
Aer2
(80)( 3.2832)
(18)(0.4499)(10)
B
3.24 ksi
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643
a
B
PROBLEM 4.185
20 mm
B
A
A
250 N · m
250 N · m
For the curved beam and loading shown, determine the stress at
(a) point A, (b) point B.
30 mm
a
35 mm
40 mm
Section a–a
SOLUTION
Locate centroid.
A, mm 2
r , mm
Ar , mm3
600
45
27 103
300
55
16.5 103
43.5 103
900
r
R
43.5 103
900
1
2
(b1r2
48.333 mm
h 2 (b1
b2r1) ln
r2
r1
b2 )
h(b2
b1)
(0.5)(30) 2 (40 20)
65
[(40)(65) (20)(35)]ln 35
(30)(40
e
(a)
yA
R
yB
B
R
R
1.4725 mm
M
46.8608 mm
250 N m
11.8608 mm
My A
Aer1
A
(b)
r1
r
20)
r2
MyB
Aer2
( 250)(11.8608 10 3 )
(900 10 6 )(1.4725 10 3 )(35 10 3 )
63.9 106 Pa
A
63.9 MPa
18.1392 mm
( 250)( 18.1392 10 3 )
(900 10 6 )(1.4725 10 3 )(65 10 3 )
52.6 106 Pa
B
52.6 MPa
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644
PROBLEM 4.186
35 mm
25 mm
For the crane hook shown, determine the largest tensile stress in
section a-a.
60 mm
40 mm
a
a
60 mm
Section a–a
15 kN
SOLUTION
Locate centroid.
A, mm 2
r , mm
Ar , mm3
1050
60
63 103
750
80
60 103
103 103
1800
r
103 103
1800
63.333 mm
P 15 103 N
Force-couple system at centroid:
M
(15 103 )(68.333 10 3 )
Pr
1
2
R
(b1r2
1.025 103 N m
h 2 (b1 b2 )
b2 r1 ) ln
r2
r1
h(b1 b2 )
(0.5)(60)2 (35 25)
[(35)(100) (25)(40)]ln 100
(60)(35 25)
40
e
r
R
63.878 mm
4.452 mm
Maximum tensile stress occurs at point A.
yA
A
R r1
P
A
23.878 mm
My A
Aer1
15 103
1800 10
6
(1.025 103 )(23.878 10 3 )
(1800 10 6 )(4.452 10 3 )(40 10 3 )
84.7 106 Pa
A
84.7 MPa
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645
PROBLEM 4.187
Using Eq. (4.66), derive the expression for R given in Fig. 4.61 for a circular cross section.
SOLUTION
Use polar coordinate
as shown. Let w be the width as a function of
w 2c sin
r r c cos
dr
c sin
d
dA
w dr
2c 2 sin 2
d
2
dA
r
2c sin
r c cos
0
dA
r
c 2 (1 cos 2 )
d
r c cos
0
r2
2
c 2 cos 2
(r 2
r c cos
0
2
d
0
(r
2r
2(r
2(r 2
c cos ) d
2c sin
0
2
c )
r2
c2
tan
1
0) 2c(0 0) 4 r 2
2r (
2 r
r2
2
d
c2 )
0
r
dr
c cos
0
2
2
c2 )
r2
c2
c 2 tan 1
2
r c
2
0
0
c2
c2
A
c2
A
dA
r
R
1
2
2 r
c2 r
r
2
(r
r2
2
r2
c2
2
2
c )
c2
1
2
c2
1
2r
c2 r
r2
r2
c
2
c2
c2
r
r2
c2
r
r2
c2
1
r
2
r2
c2
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646
PROBLEM 4.188
Using Eq. (4.66), derive the expression for R given in Fig. 4.73 for a trapezoidal cross section.
SOLUTION
The section width w varies linearly with r.
w
w
b1
b2
b2
b1
c1
r2b1
r1b2
c0
dA
r
c0 c1r
b1 at r r1 and w b2 at r
c0 c1r1
c0 c1r2
c1 (r1 r2 )
c1h
b1 b2
h
(r2 r1 )c0 hc0
r2 b1 r1b2
h
r2 w
r2 c
c1r
0
dr
dr
r1 r
r1
r
r2
r2
c0 ln r
c1 r
r1
r2
r1
r
c0 ln 2 c1 (r2 r1 )
r1
r2 b1 r1b2 r2 b1 b2
h
ln
h
r1
h
r2 b1 r1b2 r2
ln
(b1 b2 )
h
r1
1
(b1 b2 ) h
A
2
1 2
h (b1 b2 )
A
2
R
r
dA
( r2b1 r1b2 ) ln r2 h(b1 b2 )
r
1
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647
PROBLEM 4.189
Using Equation (4.66), derive the expression for R given in Fig. 4.73 for a triangular cross section.
SOLUTION
The section width w varies linearly with r.
w
w
b
0
b
c1
dA
r
c0 c1r
b at r r1 and w 0 at r r2
c0 c1r1
c0 c1r2
c1 (r1 r2 )
c1h
br2
b
and c0
c1r2
h
h
r2 w
r2 c
c1r
0
dr
dr
r1 r
r1
r
r2
r2
c0 ln r
c1 r
r1
r1
r
c0 ln 2 c1 (r2 r1 )
r1
br2 r2 b
h
ln
h
r1 h
A
R
br2 r2
ln
h
r1
1
bh
2
A
dA
r
b
b b
1
2
r2
h
r2 r2
ln
h r1
bh
ln
r2
r1
1
r2
h
1
1
2
h
ln
r2
r1
1
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648
PROBLEM 4.190
b2
b3
Show that if the cross section of a curved beam consists of two or more
rectangles, the radius R of the neutral surface can be expressed as
b1
A
R
r1
r2
r1
ln
r2
b1
r3
r2
b2
r4
r3
b3
r3
r4
where A is the total area of the cross section.
SOLUTION
A
1 dA
r
R
A
ri 1
ri
bi ln
A
ln
Note that for each rectangle,
1
dA
r
ri
1
ri
bi
ri
r2
A
ri 1
ri
bi
1
bi
r2
r1
ln
b1
r3
r2
b2
r4
r3
b3
dr
r
dr
r
bi ln
ri
1
ri
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649
PROBLEM 4.191
C
"
2
"
2
r1
!x
!x
For a curved bar of rectangular cross section subjected to a
bending couple M, show that the radial stress at the neutral
surface is
r
M
R
1 1 ln
r
Ae
R
r1
R
b
and compute the value of r for the curved bar of Concept
Applications 4.10 and 4.11. (Hint: consider the free-body
diagram of the portion of the beam located above the neutral
surface.)
!r
!r
SOLUTION
M (r R ) M
Aer
Ae
For portion above the neutral axis, the resultant force is
At radial distance r,
r
H
R
r dA
r b dr
r1
Mb R
MRb R dr
dr
Ae r1
Ae r1 r
Mb
MRb R
( R r1 )
ln
Ae
Ae
r1
Resultant of
n:
Fr
/2
r
/2
r
For equilibrium: Fr
r
cos
/2
2 H sin
2 r bR sin
2
/2
2
r
MbR
1 1
Ae
R
ln
R
r1
dA
cos (bR d )
bR sin
MR
Aer
/2
r bR
r
/2
bR sin
cos d
2
0
2
2
r
MbR
1 1
Ae
R
ln
R
sin
r1
2
r
r
M
1 1
Ae
R
ln
R
r1
0
Using results of Examples 4.10 and 4.11 as data,
M
r
8 kip in.,
A
3.75 in 2 , R
8
1
(3.75)(0.0314)
5.25
5.9686
5.9686 in., e
ln
5.9686
5.25
0.0314 in., r1
5.25 in.
r
0.536 ksi
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650
PROBLEM 4.192
25 mm
25 mm
4 kN
A
Two vertical forces are applied to a beam of the cross section shown. Determine
the maximum tensile and compressive stresses in portion BC of the beam.
4 kN
B
C
300 mm
300 mm
SOLUTION
A1
2
A2
A2 y2
A2
r2
bh
2
(25)2
981.7 mm 2
1250 mm 2
(50)(25)
y
A1 y1
A1
(981.7)(10.610) (1250)( 12.5)
981.7 1250
I1
I x1
d1
y1
r 4 A1 y12
(25)4 (981.7)(10.610) 2
8
8
y 10.610 ( 2.334) 12.944 mm
I1
I1
A1d12
(981.7)(12.944)2
d2
1 3 1
bh
(50)(25)3 65.104 103 mm 4
12
12
y2 y
12.5 ( 2.334) 10.166 mm
I2
I2
A2 d 22
I
I1
I2
I2
ytop
ybot
65.104 103
(1250)(10.166) 2
401.16 103 mm 4
401.16 10
25 2.334
25 2.334
27.334 mm
bot
Mytop
I
Mybot
I
(4)(25)
10.610 mm
3
25
12.5 mm
2
42.886 106 mm 4
207.35 103 mm 4
194.288 103 mm 4
m4
0.027334 m
22.666 mm
0.022666 m
M
top
9
y2
4r
3
h
2
2.334 mm
A1 y12
42.866 103
y1
Pa
0:
M
(1200)(0.027334)
401.16 10 9
81.76 106 Pa
(1200)( 0.022666)
401.16 10 9
67.80 106 Pa
Pa
(4 103 )(300 10 3 )
1200 N m
top
bot
81.8 MPa
67.8 MPa
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651
PROBLEM 4.193
A steel band saw blade that was originally straight passes over 8-in.-diameter pulleys
when mounted on a band saw. Determine the maximum stress in the blade, knowing that
it is 0.018 in. thick and 0.625 in. wide. Use E 29 106 psi.
0.018 in.
SOLUTION
Band blade thickness:
t
0.018 in.
Radius of pulley:
r
1
d
2
4.000 in.
Radius of curvature of centerline of blade:
r
c
1
t
2
1
t
2
4.009 in.
0.009 in.
c
0.009
4.009
Maximum strain:
m
Maximum stress:
m
E
m
65.1 103 psi
m
0.002245
(29 106 )(0.002245)
m
65.1 ksi
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652
PROBLEM 4.194
M
A couple of magnitude M is applied to a square bar of side a. For
each of the orientations shown, determine the maximum stress
and the curvature of the bar.
M
a
(a)
(b)
SOLUTION
1 3
bh
12
a
2
I
c
1
a4
12
Ma
2
a4
12
Mc
I
max
1 3
aa
12
M
EI
max
1
M
4
Ea
12
6M
a3
12M
Ea 4
For one triangle, the moment of inertia about its base is
1 3
bh
12
I1
1
12
2a
a
3
2
a4
24
4
c
I2
I1
a
24
I
I1
I2
a4
12
a
2
max
1
Mc
I
M
EI
Ma/ 2
a 4 /12
6 2M
a3
8.49M
a3
max
1
M
4
Ea
12
12M
Ea 4
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653
40 mm
60 mm
PROBLEM 4.195
Determine the plastic moment M p of a steel beam of the cross section shown, assuming
the steel to be elastoplastic with a yield strength of 240 MPa.
SOLUTION
Let c1 be the outer radius and c2 the inner radius.
A1 y1
Aa ya
2
4c1
3
c12
2 3
c1
3
c2
A1 y1
2 3
c1
3
A2 y2
Y ( A1 y1
Mp
Data:
Ab yb
2
3
c 22
A2 y2 )
4
3
Y
c13
c32
240 106 Pa
Y
240 MPa
c1
60 mm
0.060 m
c2
40 mm
0.040 m
Mp
4c2
3
c22
4
(240 106 )(0.0603
3
48.64 103 N m
0.0403 )
Mp
48.6 kN m
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654
PROBLEM 4.196
M # 300 N · m
In order to increase corrosion resistance, a 2-mm-thick cladding of
aluminum has been added to a steel bar as shown. The modulus
of elasticity is 200 GPa for steel and 70 GPa for aluminum. For a
bending moment of 300 N m, determine (a) the maximum stress
in the steel, (b) the maximum stress in the aluminum, (c) the radius
of curvature of the bar.
26 mm
30 mm
46 mm
50 mm
SOLUTION
Use aluminum as the reference material.
n
1 in aluminum
Es
Ea
n
200
70
2.857 in steel
Cross section geometry:
Steel: As
(46 mm)(26 mm)
Aluminum: Aa
1196 mm 2
1196 mm 2
(50 mm)(30 mm)
1
(50 mm)(30 mm3 )
12
Ia
1
(46 mm)(26 mm)3
12
Is
67,375 mm 4
304 mm 2
67,375 mm 4
45,125 mm 4
Transformed section.
I
na I a
ns I s
(1)(45,125)
M
300 N m
Bending moment.
(a)
ns Mys
I
s
(b)
(c)
(2.857)(300)(0.013)
237.615 10 9
Maximum stress in aluminum:
a
2.857
ns
Maximum stress in steel:
na Mya
I
(1)(300)(0.015)
237.615 10 9
Radius of curvative:
EI
M
237, 615 mm 4
(2.857)(67,375)
na
ys
13 mm
237.615 10 9 m 4
0.013 m
46.9 106 Pa
1,
ya
s
15 mm
46.9 MPa
0.015 m
18.94 106 Pa
(70 109 )(237.615 10 9 )
300
a
18.94 MPa
55.4 m
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on a website, in whole or part.
655
PROBLEM 4.197
The vertical portion of the press shown consists of a rectangular
tube of wall thickness t 10 mm. Knowing that the press has
been tightened on wooden planks being glued together until P
20 kN, determine the stress at (a) point A, (b) point B.
t
P
a
P'
a
t
A
200 mm
80 mm
60 mm
80 mm
B
Section a–a
SOLUTION
Rectangular cutout is 60 mm
40 mm.
A
(80)(60)
I
1
(60)(80)3
12
2.4 103 mm 2
(60)(40)
1
(40)(60)3
12
2.4 10 3 m 2
1.84 106 mm 4
1.84 10 6 m 4
c
P
M
(a)
(b)
40 mm
0.040 m e
200
40
240 mm
0.240 m
3
20 10 N
Pe
(20 103 )(0.240)
4.8 103 N m
A
P
A
Mc
I
20 103
2.4 10 3
(4.8 103 )(0.040)
1.84 10 6
112.7 106 Pa
A
112.7 MPa
B
P
A
Mc
I
20 103
2.4 10 3
(4.8 103 )(0.040)
1.84 10 6
96.0 106 Pa
B
96.0 MPa
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656
PROBLEM 4.198
P
P y
P
P
B
C
D
a
A
x
z
The four forces shown are applied to a rigid plate supported by a solid steel
post of radius a. Knowing that P 24 kips and a 1.6 in., determine the
maximum stress in the post when (a) the force at D is removed, (b) the forces
at C and D are removed.
SOLUTION
For a solid circular section of radius a,
Centric force:
(a)
(b)
A
a2
I
F
4P,
Mx
F
3P,
Mx
F
A
M xz
I
2P,
Mx
M
M x2
4
a4
F
A
4P
a2
( Pa)( a)
a2
4
7P
a2
0
Mz
Force at D is removed.
Pa,
Mz
3P
a2
0
Forces at C and D are removed.
F
Resultant bending couple:
F
A
2P
a2
P
Numerical data:
Answers:
Mc
I
Pa,
Mz
M z2
2 Paa
a2
4
2 Pa
2
24.0 kips,
Pa
4 2 P
a2
a
2.437 P/a 2
1.6 in.
(a)
(7)(24.0)
(1.6)2
20.9 ksi
(b)
(2.437)(24.0)
(1.6)2
22.8 ksi
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657
a
r # 20 mm
PROBLEM 4.199
P # 3 kN
The curved portion of the bar shown has an inner radius of 20 mm.
Knowing that the allowable stress in the bar is 150 MPa, determine the
largest permissible distance a from the line of action of the 3-kN force to
the vertical plane containing the center of curvature of the bar.
25 mm
25 mm
SOLUTION
Reduce the internal forces transmitted across section AB to a force-couple system at the centroid of the
section. The bending couple is
M
P(a
r)
For the rectangular section, the neutral axis for bending couple only lies at
R
Also, e
r
h
r2
r1
ln
R
The maximum compressive stress occurs at point A. It is given by
P
A
A
My A
Aer1
P
A
with
yA
R
Thus,
K
1
Data:
25 mm, r1
R
25
45
ln 20
b
25 mm,
K
a
r
a
r ) yA
Aer1
r )( R
er1
P
A
r1)
20 mm, r2
30.8288 mm, e
A
3
K
r1
(a
h
P
P (a
3 10 N m,
bh
45 mm, r
32.5
(25)(25)
A
30.8288
625 mm 2
32.5 mm
1.6712 mm
625 10 6 m 2
R
r1
10.8288 mm
6
150 10 Pa
( 150 106 )(625 10 6 )
31.25
P
3 103
( K 1)er1 (30.25)(1.6712)(20)
93.37 mm
R r1
10.8288
AA
93.37
a
32.5
60.9 mm
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658
Dimensions in inches
400 lb
400 lb
PROBLEM 4.200
400 lb
2.5
Determine the maximum stress in each of the two machine
elements shown.
400 lb
2.5
3
0.5
3
r 5 0.3
r 5 0.3
1.5
0.5
0.5
1.5
0.5
SOLUTION
For each case, M
(400)(2.5)
1000 lb in.
At the minimum section,
1
(0.5)(1.5)3
12
0.75 in.
I
c
(a)
D/d
r/d
3/1.5
0.140625 in 4
2
0.3/1.5
0.2
K
From Fig 4.32,
max
(b)
D/d
KMc
I
3/1.5
(1.75)(1000)(0.75)
0.140625
2
From Fig. 4.31,
max
1.75
KMc
I
r /d
0.3/1.5
K
9.33 103 psi
max
9.33 ksi
max
8.00 ksi
0.2
1.50
(1.50)(1000)(0.75)
0.140625
8.00 103 psi
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659
PROBLEM 4.201
120 mm
10 mm
M
120 mm
10 mm
Three 120 10-mm steel plates have been welded together to form the
beam shown. Assuming that the steel is elastoplastic with
E 200 GPa and Y 300 MPa, determine (a) the bending moment
for which the plastic zones at the top and bottom of the beam are
40 mm thick, (b) the corresponding radius of curvature of the beam.
10 mm
SOLUTION
A1
R1
A2
R2
A3
R3
y1
(a)
M
2( R1 y1
(120)(10) 1200 mm 2
Y
(300 106 )(1200 10 6 ) 360 103 N
A1
(30)(10) 300 mm 2
Y
(300 106 )(300 10 6 ) 90 103 N
A2
(30)(10)
1
Y A2
2
65 mm
R2 y2
300 mm 2
1
(300 106 )(300 10 6 ) 45 103 N
2
65 10 3 m y2 45 mm 45 10 3 m
R3 y3 )
yY
Y
E
EyY
Y
20 mm
20 10 3 m
2{(360)(65) (90)(45) (45)(20)}
56.7 103 N m
(b)
y3
M
(200 109 )(30 10 3 )
300 106
56.7 kN m
20.0 mm
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660
P
4.5 in.
PROBLEM 4.202
Q
4.5 in.
A short length of a W8 31 rolled-steel shape supports a rigid plate on which two
loads P and Q are applied as shown. The strains at two points A and B on the
centerline of the outer faces of the flanges have been measured and found to be
A
B
550 10
A
6
in./in.
B
680 10
6
in./in.
29 106 psi, determine the magnitude of each load.
Knowing that E
SOLUTION
Strains:
A
C
550 10
1
(
2
6
B
1
( 550
2
B)
A
680)10
6
615 10
6
E
A
(29 106 psi)( 550 10
6
in./in.)
15.95 ksi
C
E
C
(29 106 psi)( 615 10
6
in./in.)
17.835 ksi
W8
M
31:
(4.5 in.)( P
P
C
Q
A
;
A
9.12 in 2
S
27.5 in 3
680 10
P
A
15.95 ksi
Solve simultaneously.
Q
A
in./in.
Q)
17.835 ksi
P Q
9.12 in 2
P
At point A:
6
in./in.
A
Stresses:
At point C:
in./in.
Q
162.655 kips (1)
M
S
17.835 ksi
P
(4.5 in.)( P Q)
;
27.5 in 3
75.6 kips Q
P
Q
11.5194 kips (2)
87.1 kips
P
75.6 kips
Q
87.1 kips
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661
PROBLEM 4.203
!1
M1
A
M'1
M1
!1
M'1
B
C
!1
!1
D
Two thin strips of the same material and same cross
section are bent by couples of the same magnitude
and glued together. After the two surfaces of
contact have been securely bonded, the couples are
removed. Denoting by 1 the maximum stress and
by 1 the radius of curvature of each strip while
the couples were applied, determine (a) the final
stresses at points A, B, C, and D, (b) the final radius
of curvature.
SOLUTION
Let b
width and t
thickness of one strip.
Loading one strip, M
I1
1
1
1
M1
1 3
bt , c
12
M1c
I
M1
EI1
1
t
2
M1
bt 2
12 M 1
Et 3
After M1 is applied to each of the strips, the stresses are those given in the sketch above. They are
A
1,
1,
B
1,
C
D
1
The total bending couple is 2M1.
After the strips are glued together, this couple is removed.
M
1
b(2t )3
12
2 M 1, I
2 3
bt
3
c
t
The stresses removed are
My
I
A
2M 1 y
2 bt 3
3
3M1
bt 2
1
2
1,
3M1 y
bt 2
B
C
0,
D
3M1
bt 2
1
2
1
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662
PROBLEM 4.203 (Continued)
(a)
Final stresses:
A
1
(
1
2
1)
1
2
A
B
1
C
D
1
(b)
M
EI
2M 1
E 23 bt 3
Final radius:
3M 1
Et 3
1
1
1
2
D
1
1
1
1
2
1
4
3
1
1 1
4
1
1
1
1
1
1 1
4 1
3 1
4 1
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663
PROBLEM 4.C1
Two aluminum strips and a steel strip are to be bonded together to form
a composite member of width b 60 mm and depth h 40 mm. The
modulus of elasticity is 200 GPa for the steel and 75 GPa for the
aluminum. Knowing that M 1500 N m, write a computer program
to calculate the maximum stress in the aluminum and in the steel for
values of a from 0 to 20 mm using 2-mm increments. Using appropriate
smaller increments, determine (a) the largest stress that can occur in the
steel, (b) the corresponding value of a.
Aluminum
a
Steel
h " 40 mm
a
b " 60 mm
SOLUTION
Transformed section:
(all steel)
I
At Point 1:
At Point 2:
Esteel
Ealum
n
1 3
bh
12
h
2
M
alum
steel
1
1
2
(nb b) (h 2a )3
12
2
I
n
M
h
2
a
I
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664
PROBLEM 4.C1 (Continued)
For a
0 to 20 mm using 2-mm intervals: compute: n, I ,
b
Moduli of elasticity:
Steel
60 mm h
alum ,
40 mm M
200 GPa Aluminum
steel .
1500 N m
75 GPa
Program Output
a
mm
I
m 4 /106
Sigma
Aluminum
MPa
Sigma
Steel
MPa
0.000
0.8533
35.156
93.750
2.000
0.7088
42.325
101.580
4.000
0.5931
50.585
107.914
6.000
0.5029
59.650
111.347
8.000
0.4352
68.934
110.294
10.000
0.3867
77.586
103.448
12.000
0.3541
84.714
90.361
14.000
0.3344
89.713
71.770
16.000
0.3243
92.516
49.342
18.000
0.3205
93.594
24.958
20.000
0.3200
93.750
0.000
Find ‘a’ for max. steel stress and the corresponding aluminum stress.
6.600
0.4804
62.447
111.572083
6.610
0.4800
62.494
111.572159
6.620
0.4797
62.540
111.572113
Max. steel stress 111.6 MPa occurs when a
Corresponding aluminum stress
6.61 mm.
62.5 MPa
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665
tf
PROBLEM 4.C2
y
x
tw
d
bf
A beam of the cross section shown, made of a steel that is assumed to be
elastoplastic with a yield strength Y and a modulus of elasticity E, is bent
about the x axis. (a) Denoting by yY the half thickness of the elastic core, write
a computer program to calculate the bending moment M and the radius of
curvature
for values of yY from 12 d to 16 d using decrements equal
1
to 2 t f . Neglect the effect of fillets. (b) Use this program to solve Prob. 4.201.
SOLUTION
Compute moment of inertia I x .
1
bf d 3
12
Ix
Maximum elastic moment:
MY
Y
1
(b f
12
tw )(d
2t f )3
Ix
(d/2)
For yielding in the flanges,
(Consider upper half of cross section.)
d
2
c
Stress at junction of web and flange:
(d/2) t f
A
Y
yY
Resultant forces:
Detail of stress diagram:
a1
1
(c
2
a2
yY
a3
yY
a4
2
(c t f )
3
yY )
1
yY
3
2
[ yY
3
(c t f )
(c t f ) ]
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666
PROBLEM 4.C2 (Continued)
Bending moment.
4
2
M
Rn an
n 1
Radius of curvature.
yY
Y
Y
E
yY E
;
Y
(Consider upper half of cross section.)
For yielding in the web,
a5
a6
a7
1
tf
2
c
1
[ yY
2
2
yY
3
(c t f )]
7
Bending moment.
M
2
Rn an
n 5
Radius of curvature.
yY
Y
yY E
Y
E
Y
Program: Key in expressions for an and Rn for n 1 to 7.
For yY
c to (c tf ) at
tf /2 decrements, compute M
2 Rn an for n 1 to 4 and
yY E
, then print.
Y
For yY
(c tw ) to c /3 at
tf /2 decrements, compute M
2 Rn an for n 5 to 7 and
yY E
, then print.
Y
Input numerical values and run program.
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667
PROBLEM 4.C2 (Continued)
Program Output
For a beam of Problem 4.201,
Depth d
140.00 mm
Thickness of flange t f
I
10.00 mm
Width of flange b f
120.00 mm
Thickness of web tw
10.00 mm
0.000011600 m to the 4th
Yield strength of steel sigmaY
Yield moment M Y
300 MPa
49.71 kip in.
yY (mm)
M (kN m)
(m)
For yielding still in the flange,
70.000
49.71
46.67
65.000
52.59
43.33
60.000
54.00
40.00
For yielding in the web,
60.000
54.00
40.00
55.000
54.58
36.67
50.000
55.10
33.33
45.000
55.58
30.00
40.000
56.00
26.67
35.000
56.38
23.33
30.000
56.70
20.00
25.000
56.97
16.67
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668
PROBLEM 4.C3
y
#
0.4
0.4
A
B
z
C
1.2
0.4
#
M
1.2
D
E
0.8 0.4
1.6
An 8 kip in. couple M is applied to a beam of the cross section shown
in a plane forming an angle
with the vertical. Noting that the
centroid of the cross section is located at C and that the y and z axes
are principal axes, write a computer program to calculate the stress at
A, B, C, and D for values of
from 0 to 180° using 10° increments.
(Given: I y 6.23 in 4 and I z 1.481 in 4 . )
0.4 0.8
Dimensions in inches
SOLUTION
Input coordinates of A, B, C, D.
zA
z (1)
2
yA
y (1) 1.4
zB
z (2)
2
yB
y (2) 1.4
zC
z (3)
1
yC
y (3)
1.4
zD
z (4) 1
yD
y (4)
1.4
My
M sin
Components of M.
Mz
Equation 4.55, Page 305:
Program: For
( n)
M cos
M z y ( n)
Iz
M y z ( n)
Iy
0 to 180 using 10° increments.
For n 1 to 4 using unit increments.
Evaluate Equation 4.55 and print stresses.
Return
Return
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669
PROBLEM 4.C3 (Continued)
Program Output
Moment of couple:
M
8.00 kip in.
Moments of inertia:
Iy
6.23 in 4
Iz
1.481 in 4
Coordinates of Points A, B, D, and E:
Point A: z (1)
Point B: z (2)
2:
y (1) 1.4
2: y (2) 1.4
Point D: z (3)
1:
Po int E: z (4) 1:
Beta
y (3)
y (4)
---Stress at Points--A
B
ksi
ksi
1.4
1.4
D
ksi
E
ksi
0
–7.565
–7.565
7.565
7.565
10
–7.896
–7.004
7.673
7.227
20
–7.987
–6.230
7.548
6.669
30
–7.836
–5.267
7.193
5.909
40
–7.446
–4.144
6.621
4.970
50
–6.830
–2.895
5.846
3.879
60
–6.007
–1.558
4.895
2.670
70
–5.001
–0.174
3.794
1.381
80
–3.843
1.216
2.578
0.049
90
–2.569
2.569
1.284
–1.284
100
–1.216
3.843
–0.049
–2.578
110
0.174
5.001
–1.381
–3.794
120
1.558
6.007
–2.670
–4.895
130
2.895
6.830
–3.879
–5.846
140
4.144
7.446
–4.970
–6.621
150
5.267
7.836
–5.909
–7.193
160
6.230
7.987
–6.669
–7.548
170
7.004
7.896
–7.227
–7.673
180
7.565
7.565
–7.565
–7.565
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670
PROBLEM 4.C4
b
B
B
A
A
h
M'
M
r1
C
Couples of moment M 2 kN m are applied as shown to a curved bar
having a rectangular cross section with h 100 mm and b 25 mm. Write
a computer program and use it to calculate the stresses at points A and B for
values of the ratio r1/h from 10 to 1 using decrements of 1, and from 1
to 0.1 using decrements of 0.1. Using appropriate smaller increments,
determine the ratio r1/h for which the maximum stress in the curved bar is
50% larger than the maximum stress in a straight bar of the same cross
section.
SOLUTION
Input:
h 100 mm,
b 25 mm,
M 2 kN m
For straight bar,
straight
M
S
6M
h 2b
48 MPa
Following notation of Section 4.15, key in the following:
r2
Stresses:
A
h r1 ; R
1
M (r1
h /ln (r2
r1 ); r
R)( Aer1 )
r1
2
B
r2 : e
M (r2
r
R; A bh
R)/(Aer2 )
2500
(I)
(II)
Since h 100 mm, for r1/h 10, r1 1000 mm. Also, r1/h 10, r1 100
Program:
For r1 1000 to 100 at
100 decrements,
using equations of Lines I and II, evaluate r2 , R, r , e,
Also evaluate ratio
1/
Return and repeat for r1
1,
and
2
straight
100 to 10 at
10 decrements.
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671
PROBLEM 4.C4 (Continued)
Program Output
M
Bending moment
Stress in straight beam
2 kN m h 100.000 in. A 2500.00 mm 2
48.00 MPa
r1
mm
rbar
mm
R
mm
e
mm
MPa
MPa
r1 /h
–
ratio
–
1000
1050
1049
0.794
–49.57
46.51
10.000
–1.033
900
950
949
0.878
–49.74
46.36
9.000
–1.036
800
850
849
0.981
–49.95
46.18
8.000
–1.041
700
750
749
1.112
–50.22
45.95
7.000
–1.046
600
650
649
1.284
–50.59
45.64
6.000
–1.054
500
550
548
1.518
–51.08
45.24
5.000
–1.064
400
450
448
1.858
–51.82
44.66
4.000
–1.080
300
350
348
2.394
–53.03
43.77
3.000
–1.105
200
250
247
3.370
–55.35
42.24
2.000
–1.153
100
150
144
5.730
–61.80
38.90
1.000
–1.288
1
2
=====================================================
100
150
144
5.730
–61.80
38.90
1.000
–1.288
90
140
134
6.170
–63.15
38.33
0.900
–1.316
80
130
123
6.685
–64.80
37.69
0.800
–1.350
70
120
113
7.299
–66.86
36.94
0.700
–1.393
60
110
102
8.045
–69.53
36.07
0.600
–1.449
50
100
91
8.976
–73.13
35.04
0.500
–1.523
40
90
80
10.176
–78.27
33.79
0.400
–1.631
30
80
68
11.803
–86.30
32.22
0.300
–1.798
20
70
56
14.189
–100.95
30.16
0.200
–2.103
10
60
42
18.297
–138.62
27.15
0.100
–2.888
===================================================================
Find r1/h for (
max ) /( straight )
1.5
52.70
52.80
103
103
94
94
8.703
8.693
–72.036
–71.998
35.34
35.35
0.527
0.528
–1.501
–1.500
52.90
103
94
8.683
–71.959
35.36
0.529
–1.499
Ratio of stresses is 1.5 for r1
52.8 mm or r1/h
0.529.
[Note: The desired ratio r1/h is valid for any beam having a rectangular cross section.]
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672
bn
hn
h2
PROBLEM 4.C5
M
The couple M is applied to a beam of the cross
section shown. (a) Write a computer program that,
for loads expressed in either SI or U.S. customary
units, can be used to calculate the maximum tensile
and compressive stresses in the beam. (b) Use this
program to solve Probs. 4.9, 4.10, and 4.11.
b2
h1
b1
SOLUTION
Input:
Bending moment M.
For n 1 to n,
Enter bn and hn
bn hn
Area
an
an
(Print)
(hn 1 )/2 hn /2
1
[Moment of rectangle about base]
m
( Area)an
m
m
[For whole cross section]
m; Area
Area
Area
Location of centroid above base.
y
m/Area
(Print)
Moment of inertia about horizontal centroidal axis.
For n 1 to n,
an
an
(hn 1 )/2 hn /2
1
I
bn hn3 /12 (bn hn )( y
I
I
I
an )2
(Print)
Computation of stresses.
For n 1 to n,
Total height:
H
H
hn
Stress at top:
M top
M
H
y
I
(Print)
Stress at bottom:
M bottom
M
y
I
(Print)
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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673
PROBLEM 4.C5 (Continued)
Problem 4.9
Summary of cross section dimensions:
Width (in.)
Height (in.)
9.00
2.00
3.00
6.00
Bending moment 600.000 kip in.
Centroid is 3.000 mm above lower edge.
Centroidal moment of inertia is 204.000 in4.
Stress at top of beam
14.706 ksi
Stress at bottom of beam 8.824 ksi
Problem 4.10
Summary of cross section dimensions:
Width (in.)
Height (in.)
4.00
1.00
1.00
6.00
8.00
1.00
Bending moment 500.000 kip in.
Centroid is 4.778 in. above lower edge.
Centroidal moment of inertia is 155.111 in4.
Stress at top of beam
10.387 ksi
Stress at bottom of beam 15.401 ksi
Problem 4.11
Summary of cross section dimensions:
Width (mm)
Height (mm)
50
10
20
50
Bending moment 1500.0000 N m
Centroid is 25.000 mm above lower edge.
Centroidal moment of inertia is 512,500 mm4.
Stress at top of beam
102.439 MPa
Stress at bottom of beam 72.171 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
674
PROBLEM 4.C6
y
Dy
y
c
M
z
A solid rod of radius c 1.2 in. is made of a steel that is assumed to be
elastoplastic with E 29,000 ksi and Y 42 ksi. The rod is subjected to a
couple of moment M that increases from zero to the maximum elastic
moment M Y and then to the plastic moment M p . Denoting by yY the half
thickness of the elastic core, write a computer program and use it to calculate
the bending moment M and the radius of curvature
for values of yY from
1.2 in. to 0 using 0.2-in. decrements. (Hint: Divide the cross section into 80
horizontal elements of 0.03-in. height.)
SOLUTION
MY
Y
Mp
Y
c3 (42 ksi) (1.2 in.)3 57 kip in.
4
4
4 3
4
c (42 ksi) (1.2 in.)3 96.8 kip in.
3
3
Consider top half of rod.
Let
i
Number of elements in top half.
c
Height of each element: h
L
h
For n
0 to i 1, Step 1:
y
n( h)
c 2 {(n 0.5) h}2
z
If y
z at midheight of element
yY go to 100
(n 0.5) h
E
Y
Stress in elastic core
y
go to 200
100
E
200
Area
Force
Moment
M
P
Y
Stress in plastic zone
2 z ( h)
E ( Area)
Repeat for yY 1.2 in.
to yY 0
At 0.2-in. decrements
Force (n 0.5) h
Moment
M
yY E / Y
Print yY , M , and .
Next
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
675
PROBLEM 4.C6 (Continued)
Program Output
Radius of rod 1.2 in.
Yield point of steel
Yield moment
Plastic moment
Number of elements in half of the rod
42 ksi
57.0 kip in.
96.8 kip in.
40
For yY
1.20 in.,
M
57.1 kip in.
Radius of curvature
828.57 in.
For yY
1.00 in.,
M
67.2 kip in.
Radius of curvature
690.48 in.
For yY
0.80 in.,
M
76.9 kip in.
Radius of curvature
552.38 in.
For yY
0.60 in.,
M
85.2 kip in.
Radius of curvature
414.29 in.
For yY
0.40 in.,
M
91.6 kip in.
Radius of curvature
276.19 in.
For yY
0.20 in.,
M
95.5 kip in.
Radius of curvature
138.10 in.
For yY
0.00 in.,
M
infinite
Radius of curvature
zero
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
676
PROBLEM 4.C7
2 in.
C
3 in.
The machine element of Prob 4.178 is to be redesigned by removing part
of the triangular cross section. It is believed that the removal of a small
triangular area of width a will lower the maximum stress in the element.
In order to verify this design concept, write a computer program to
calculate the maximum stress in the element for values of a from 0 to
1 in. using 0.1-in. increments. Using appropriate smaller increments,
determine the distance a for which the maximum stress is as small as
possible and the corresponding value of the maximum stress.
B
A
2.5 in.
a
SOLUTION
See Figure 4.79, Page 289.
M
For a
5 kip in. r2
5 in. b2
2.5 in.
0 to 1.0 at 0.1 intervals,
h 3 a
r1 2 a
b1
b2 (a /(h a ))
Area
(b1 b2 )(h/2)
x
a+
r
r2
R
e
1
b1h(h/3)
2
Area
(h x )
1
2
(b1r2
r
1
b2 h 2h/3
2
h 2 (b1 b2 )
b2 r1 ) ln
r2
r1
h(b1 b2 )
R
D
M (r1
R)/[Area (e)(r1 )]
B
M (r2
R) /[Area (e)(r2 )]
Print and Return
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
677
PROBLEM 4.C7 (Continued)
Program Output
a
in.
R
in.
D
B
ksi
ksi
b1
r
e
0.00
3.855
8.5071
2.1014
0.00
4.00
0.145
0.10
3.858
7.7736
2.1197
0.08
4.00
0.144
0.20
3.869
7.2700
2.1689
0.17
4.01
0.140
0.30
3.884
6.9260
2.2438
0.25
4.02
0.134
0.40
3.904
6.7004
2.3423
0.33
4.03
0.127
0.50
3.928
6.5683
2.4641
0.42
4.05
0.119
0.60
3.956
6.5143
2.6102
0.50
4.07
0.111
0.70
3.985
6.5296
2.7828
0.58
4.09
0.103
0.80
4.018
6.6098
2.9852
0.67
4.11
0.094
0.90
4.052
6.7541
3.2220
0.75
4.14
0.086
1.00
4.089
6.9647
3.4992
0.83
4.17
0.078
Determination of the maximum compressive stress that is as small as possible.
a
in.
R
in.
D
B
ksi
ksi
b1
r
e
0.620
3.961
6.51198
2.6425
0.52
4.07
0.109
0.625
3.963
6.51185
2.6507
0.52
4.07
0.109
0.630
3.964
6.51188
2.6591
0.52
4.07
0.109
Answer: When a
625 in., the compressive stress is 6.51 ksi.
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
678